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Section 2.8 Combining Functions; Composite Functions

Before Starting this Section, Review

1 Domain of a function (Section 2.4 , page 216)
2 Rational inequalities (Section 1.6 , page 152)

Objectives

1 Define basic operations on functions.
2 Form composite functions.
3 Find the domain of a composite function.
4 Decompose a function.
5 Apply composition to practical problems.

BP Oil Spill

The Deepwater Horizon was a floating drilling rig that could operate in waters up to 10,000 feet deep. British Petroleum (BP) was the operator and principal developer of the exploratory well situated in the Gulf of Mexico about 41 miles off the Louisiana coast. On April 20, 2010, at approximately 9:45 p.m., high-pressure methane gas from the well ignited and exploded, engulfing the platform of the rig. At the time, 126 crew members were on board. Of these, 11 workers were never found despite extensive search operations and are believed to have died in the explosion.

The total estimated volume of oil leaked was approximately 5 million barrels, making it the largest accidental oil spill in history. The spill directly impacted 68,000 square miles (about the size of Oklahoma). By June 2010, oil had washed up on 1074 miles of the coasts of Louisiana, Mississippi, Alabama, and Florida.

On June 16, 2010, BP announced a $20 billion fund to settle claims arising from the disaster. The fund was to be used only for natural resource restoration and compensation for individual and business claimants. In July, 2013, BP and its partners in the oil well, Transocean and Halliburton, went on trial to determine payouts and fines under the Clean Water Act and the Natural Resources Damages Assessment. In July, 2015 BP said its total pre-tax charge for the spill stood at 53.8 billion.

In Example 9, we calculate the area covered by an oil spill.

Combining Functions

1 Define basic operations on functions.

Just as numbers can be added, subtracted, multiplied, and divided to produce new numbers, so functions can be added, subtracted, multiplied, and divided to produce other functions. To add, subtract, multiply, or divide functions, we simply add, subtract, multiply, or divide their output values.

For example, if $f (x) = x^{2} - 2 x + 4$ and $g (x) = x + 2,$ then

$f (x) + g (x) = (x^{2} - 2 x + 4) + (x + 2) = x^{2} - x + 6 .$

This new function $y = x^{2} - x + 6$ is called the sum function $f + g .$

Example 1 Combining Functions

Let $f (x) = x^{2} - 6 x + 8$ and $g (x) = x - 2 .$ Find each of the following functions or function values.

$(f - g) (4)$
$(\frac{f}{g}) (3)$
$(f + g) (x)$
$(f - g) (x)$
(fg)(x)
$(\frac{f}{g}) (x)$

Solution

$\begin{array}{rcll} f (4) & = & 4^{2} - 6 (4) + 8 = 0 & Replace x with 4 in f (x) . \\ g (4) & = & 4 - 2 = 2 & Replace x with 4 in g (x) . \\ (f - g) (4) & = & f (4) - g (4) & Definition of difference \\ = & 0 - 2 = - 2 & Replace f (4) with 0, g (4) with 2. \end{array}$
$\begin{array}{rcll} f (3) & = & 3^{2} - 6 (3) + 8 = - 1 & Replace x with 3 in f (x) . \\ g (3) & = & 3 - 2 = 1 & Replace x with 3 in g (x) . \\ (\frac{f}{g}) (3) & = & \frac{f (3)}{g (3)} = \frac{- 1}{1} = - 1 & Replace f (3) with - 1, g (3) with 1. \end{array}$
$\begin{array}{rcll} (f + g) (x) & = & f (x) + g (x) & Definition of sum \\ = & (x^{2} - 6 x + 8) + (x - 2) & Add f (x) and g (x) . \\ = & x^{2} - 5 x + 6 & Combine like terms . \end{array}$
$\begin{array}{rcll} (f - g) (x) & = & f (x) - g (x) & Definition of difference \\ = & (x^{2} - 6 x + 8) - (x - 2) & Subtract g (x) from f (x) . \\ = & x^{2} - 7 x + 10 & Combine like terms . \end{array}$
$\begin{array}{rcll} (f g) (x) & = & f (x) \cdot g (x) & Definition of product \\ = & (x^{2} - 6 x + 8) (x - 2) & Multiply f (x) and g (x) . \\ = & x^{2} (x - 2) - 6 x (x - 2) + 8 (x - 2) & Distributive property \\ = & x^{3} - 2 x^{2} - 6 x^{2} + 12 x + 8 x - 16 & Distributive property \\ = & x^{3} - 8 x^{2} + 20 x - 16 & Combine like terms . \end{array}$
$\begin{array}{rcll} (\frac{f}{g}) (x) & = & \frac{f (x)}{g (x)}, g (x) \neq 0 & Definition of quotient \\ = & \frac{x^{2} - 6 x + 8}{x - 2}, x - 2 \neq 0 & \begin{array}{l} Replace f (x) with x^{2} - 6 x + 8, g (x) \\ with x - 2. \end{array} \\ = & \frac{(x - 2) (x - 4)}{x - 2}, x \neq 2 & Factor the numerator . \\ = & x - 4, x \neq 2 \end{array}$

Because f and g are polynomials, the domain of both f and g is the set of all real numbers, or in interval notation, $(- \infty, \infty) .$ So the domain for $f + g, f - g,$ and fg is $(- \infty, \infty) .$ However, for $\frac{f}{g},$ we must exclude 2 because $g (2) = 0 .$ The domain for $\frac{f}{g}$ is the set of all real numbers $x, x \neq 2,$ or in interval notation, $(- \infty, 2) \cup (2, \infty) .$

Practice Problem 1

Let $f (x) = 3 x - 1$ and $g (x) = x^{2} + 2 .$ Find $(f + g) (x), (f - g) (x), (f g) (x),$ and $(\frac{f}{g}) (x) .$

Example 2 Finding the Domain

Let $f (x) = \sqrt{x + 2}$ and $g (x) = \sqrt{5 - x} .$ Find the domain of $f + g, f - g, f g,$ and $\frac{f}{g} .$

Solution

Recall (see page 61) that $\sqrt{u}$ is defined only if $u \geq 0 .$

$\begin{array}{l} So f (x) = \sqrt{x + 2} is defined if x + 2 \geq 0 or x \geq - 2. & Solve for x . \\ The domain of f = D_{1} = [- 2, \infty) . & Interval notation \end{array}$

Similarly, $g (x) = \sqrt{5 - x}$ is defined only if $5 - x \geq 0$

$\begin{aligned} or & 5 \geq x & Solve for x . \\ or & x \leq 5. & Rewrite . \\ The domain of g = D_{2} = (- \infty, 5] . & Interval notation \end{aligned}$

To find the required domains, we first find $D_{1} \cap D_{2}$ from the graphs of $D_{1}$ and $D_{2}$ . See Figure 2.97.

The domain of $f + g, f - g,$ and fg is $D_{1} \cap D_{2} = [- 2, 5] .$ For the domain of $\frac{f}{g},$ we must exclude from $D_{1} \cap D_{2}$ those numbers where $g (x) = \sqrt{5 - x} = 0 .$ Solving this equation for x, we have $x = 5 .$

So the domain of $\frac{f}{g}$ is $[- 2, 5) .$

Practice Problem 2

Let $f (x) = \sqrt{x - 1}$ and $g (x) = \sqrt{3 - x} .$ Find the domain of $f g, \frac{f}{g},$ and $\frac{g}{f} .$

Composition of Functions

2 Form composite functions.

There is yet another way to construct a new function from two functions. Figure 2.98 shows what happens when you apply one function $g (x) = x^{2} - 1$ to an input (domain) value, x, and then apply a second function $f (x) = \sqrt{x}$ to the output value, g(x), from the first function. In this case, the output from the first function becomes the input for the second function.

Figure 2.99 may help clarify the definition of $f \circ g .$

Figure 2.99 Composition of the function `f` with `g`

Evaluating $(f \circ g) (x)$

By definition,

Thus, to evaluate $(f \circ g) (x),$ we must

Evaluate the inner function g at x.
Use the result g(x) as the input for the outer function f.

Example 3 Evaluating a Composite Function

Let $f (x) = x^{3}$ and $g (x) = x + 1 .$ Find each composite function value.

$(f \circ g) (1)$
$(g \circ f) (1)$
$(f \circ f) (- 1)$

Solution

$\begin{array}{rcll} (f \circ g) (1) & = & f (g (1)) & Definition of f \circ g \\ = & {[g (1)]}^{3} & Replace x with g (1) in f (x) = x^{3} . \\ = & 2^{3} = 8 & g (1) = 1 + 1 = 2; simplify . \end{array}$
$\begin{array}{rcll} (g \circ f) (1) & = & g (f (1)) & Definition of g \circ f \\ = & f (1) + 1 & Replace x with f (1) in g (x) = x + 1. \\ = & 1 + 1 = 2 & f (1) = 1^{3} = 1; simplify . \end{array}$
$\begin{array}{rcll} (f \circ f) (- 1) & = & f (f (- 1)) & Definition of f \circ f \\ = & {[f (- 1)]}^{3} & Replace x with f (- 1) in f (x) = x^{3} . \\ = & {(- 1)}^{3} = - 1 & f (- 1) = {(- 1)}^{3} = - 1; simplify . \end{array}$

Practice Problem 3

Let $f (x) = - 5 x$ and $g (x) = x^{2} + 1 .$ Find each composite function value.
1. $(f \circ g) (0)$
2. $(g \circ f) (0)$

Side Note

When computing $(f \circ g) (x) = f (g (x))$ , you must replace every x in f(x) with g(x).

Example 4 Finding Composite Functions

Let $f (x) = 2 x + 1$ and $g (x) = x^{2} - 3 .$ Find each composite function.

$(f \circ g) (x)$
$(g \circ f) (x)$
$(f \circ f) (x)$

Solution

$\begin{array}{rcll} (f \circ g) (x) & = & f (g (x)) & Definition of f \circ g \\ = & 2 g (x) + 1 & Replace x with g (x) in f (x) = 2 x + 1. \\ = & 2 (x^{2} - 3) + 1 & Replace g (x) with x^{2} - 3. \\ = & 2 x^{2} - 5 & Simplify . \end{array}$
$\begin{array}{rcll} (g \circ f) (x) & = & g (f (x)) & Definition of g \circ f \\ = & {[f (x)]}^{2} - 3 & Replace x with f (x) in g (x) = x^{2} - 3. \\ = & {(2 x + 1)}^{2} - 3 & Replace f (x) with 2 x + 1. \\ = & 4 x^{2} + 4 x + 1 - 3 & {(A + B)}^{2} = A^{2} + 2 A B + B^{2} \\ = & 4 x^{2} + 4 x - 2 & Simplify . \end{array}$
$\begin{array}{rcll} (f \circ f) (x) & = & f (f (x)) & Definition of f \circ f \\ = & 2 f (x) + 1 & Replace x with f (x) in f (x) = 2 x + 1. \\ = & 2 (2 x + 1) + 1 & Replace f (x) with 2 x + 1. \\ = & 4 x + 3 & Simplify . \end{array}$

Practice Problem 4

Let $f (x) = 2 - x$ and $g (x) = 2 x^{2} + 1 .$ Find each composite function.
1. $(g \circ f) (x)$
2. $(f \circ g) (x)$
3. $(g \circ g) (x)$

Parts a and b of Example 3 illustrate that in general, $f \circ g \neq g \circ f .$ In other words, the composition of functions is not commutative.

Domain of Composite Functions

3 Find the domain of a composite function.

Let f and g be two functions. The domain of the composite function $f \circ g$ consists of those values of x in the domain of g for which g(x) is in the domain of f. So to find the domain of $f \circ g,$ we first find the domain of g. Then from the domain of g, we exclude those values of x for which g(x) is not in the domain of f.

Procedure in Action

Example 5 Finding the Domain of a Composite Function

OBJECTIVE	EXAMPLE
Find the domain of $f \circ g$ from the defining equations for `f` and `g`.	Find the domain of $f \circ g$ if $f (x) = \frac{2}{x - 3} and g (x) = \frac{5}{x + 1} .$
Step 1 Let `A` be the domain of the inner function `g`. We may find `A` by first finding the values of `x` for which `g` is not defined.	$g (x) = \frac{5}{x + 1}$ is not defined if the denominator $x + 1 = 0$ , or $x = - 1 .$ So $A = {x \| x \neq - 1 \|}$ , or $A = (- \infty, - 1) \cup (- 1, \infty) .$
Step 2 Let `B` be the set of values of `x` for which `g`(`x`) is in the domain of `f`. As before, we may do this by finding the values of `x` for which `g`(`x`) is not in the domain of `f`.	$f (g (x)) = \frac{2}{g (x) - 3},$ so `f`(`g`(`x`)) is not defined if the denominator $\begin{array}{rcll} g (x) - 3 & = & 0 . \\ \frac{5}{x + 1} - 3 & = & 0 & Replace g (x) with \frac{5}{x + 1} . \\ 5 - 3 (x + 1) & = & 0 & Multiply by (x + 1) . \\ 5 - 3 x - 3 & = & 0 & Simplify . \\ x & = & \frac{2}{3} & Solve for x . \end{array}$ $B = {x \| x \neq \frac{2}{3}}, or B = (- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty)$
Step 3 The domain of $f \circ g$ is $A \cap B .$ The set $A \cap B$ is the values of `x` that are in the domain of `g` and for which `g`(`x`) is in the domain of `f`.	$A \cap B = (- \infty, - 1) \cup (- 1, \frac{2}{3}) \cup (\frac{2}{3}, \infty)$ is the domain of $f \circ g .$

Practice Problem 5

Let $f (x) = \sqrt{x + 1}$ and $g (x) = \frac{2}{x - 3} .$ Find the domain of $f \circ g .$

Sometimes we can determine the domain of $f \circ g$ by just examining the equation for f(g(x)).

Example 6 Finding the Domain of Composite Functions

Let $f (x) = \frac{1}{x}$ and $g (x) = 3 x - 12$ . Find the domain of $f \circ g$ .

Solution

Because $f (g (x)) = \frac{1}{g (x)} = \frac{1}{3 x - 12}$ we see f(g(x)) is not defined if the denominator is zero.

$\begin{array}{rcll} 3 x - 12 & = & 0 & Set g (x) = 0 . \\ 3 x & = & 12 & Add 12 to both sides . \\ x & = & 4 & Divide both sides by 3 . \end{array}$

So $f (g (x)) = \frac{1}{3 x - 12}$ is defined if $x \neq 4$ . The domain of $f \circ g$ is $(- \infty, 4) \cup (4, \infty)$ .

Practice Problem 6

Let $f (x) = \sqrt{x}$ and $g (x) = 3 - x$ . Find the domain of $f \circ g$ .

Side Note

If the function $f \circ g$ can be simplified, determine the domain before simplifying. For example, if $f (x) = x^{2}$ and $g (x) = \sqrt{x}$ ,

$(f \circ g) (x) = f (g (x)) = f (\sqrt{x}) .$

Before simplifying:

$(f \circ g) (x) = {(\sqrt{x})}^{2}, x \geq 0 .$

The domain of f is $(- \infty, \infty)$ , and $(f \circ g) (x)$ is defined on $[0, \infty) .$

After simplifying:

$(f \circ g) (x) = x .$

Here it appears that the domain of $f \circ g$ is $(- \infty, \infty) .$ However, this is not correct; it is $[0, \infty) .$

Example 7 Finding the Domain of Composite Functions

Let $f (x) = \sqrt{x - 2}$ and $g (x) = \sqrt{4 - x}$ . Find the following functions and their domains, using the procedure given in Example 5.

$f \circ g$
$g \circ f$

Solution

1. $g (x) = \sqrt{4 - x}$ is defined if $4 - x \geq 0$ or $x \leq 4$ . So $A = (- \infty, 4] .$
2. $\begin{array}{l} f (g (x)) = \sqrt{g (x) - 2} & Replace x with g (x) in f (x) = \sqrt{x - 2} . \end{array}$
  
  So, g(x) is in the domain of f if:
  
  $\begin{array}{rcll} g (x) - 2 & \geq & 0 \\ g (x) & \geq & 2 & Add 2 to both sides . \\ \sqrt{4 - x} & \geq & 2 & Replace g (x) with \sqrt{4 - x} . \\ 4 - x & \geq & 4 & Square both sides . \\ - x & \geq & 0 & Subtract 4 from both sides . \\ x & \geq & 0 & Solve for x . \end{array}$
  
  So $B = (- \infty, 0] .$
3. $\begin{array}{rcll} A \cap B & = & (- \infty, 4] \cap (- \infty, 0] \\ = & (- \infty, 0] See figure . \end{array}$
  
  So, the domain of $f \circ g$ is $(- \infty, 0]$
To find the domain of $g \circ f$ , we interchange the role of g and f.
1. $f (x) = \sqrt{x - 2}$ is defined if $x - 2 \geq 0$ or $x \geq 2$ . So $A = [2, \infty) .$
2. $\begin{array}{ll} g (f (x)) = \sqrt{4 - f (x)} & Replace x with f (x) in g (x) = \sqrt{4 - x} . \end{array}$
  
  So, f(x) is in the domain of g if:
  
  $\begin{array}{rcll} 4 - f (x) & \geq & 0 \\ 4 & \geq & f (x) & Add f (x) to both sides . \\ f (x) & \leq & 4 & Rewrite \\ \sqrt{x - 2} & \leq & 4 & Replace f (x) with \sqrt{x - 2} . \\ x - 2 & \leq & 16 & Square both sides (if 0 < a \leq b then a^{2} \leq b^{2}) . \\ x & \leq & 18 & Solve for x . \end{array}$
  
  So $B = (- \infty, 18] .$
3. $\begin{array}{l} A \cap B & = & [2, \infty) \cap (- \infty, 18] \\ = & [2, 18] See figure . \end{array}$
  
  So, the domain of $g \circ f = [2, 18] .$

Practice Problem 7

Let $f (x) = \sqrt{x - 1}$ and $g (x) = \sqrt{4 - x^{2}} .$ Find the domain of each of the following functions.
1. $f \circ g$
2. $g \circ f$

Decomposition of a Function

4 Decompose a function.

In the composition of two functions, we combine two functions and create a new function. However, sometimes it is more useful to use the concept of composition to decompose a function into simpler functions. For example, consider the function $H (x) = \sqrt{x^{2} - 2} .$ A way to write H(x) as the composition of two functions is to let $f (x) = \sqrt{x}$ and $g (x) = x^{2} - 2$ so that $f (g (x)) = f (x^{2} - 2) = \sqrt{x^{2} - 2} = H (x) .$

A function may be decomposed into simpler functions in several different ways by choosing various “inner” functions.

Procedure in Action

Example 8 Decomposing a Function

OBJECTIVE	EXAMPLE
Write a function H as a composite of simpler functions `f` and `g` so that $H = f \circ g$ .	Write $H (x) = \frac{1}{\sqrt{2 x^{2} + 1}}$ as $f \circ g$ for some functions `f` and `g`.
Step 1 Define `g`(`x`) as any expression in the defining equation for `H` that contains every occurrance of the variable `x`.	Let $g (x) = 2 x^{2} + 1 .$
Step 2 To get `f`(`x`) from the defining equation for `H`, (a) replace the letter `H` with `f` and (b) replace the expression chosen for `g`(`x`) with `x`.	$H (x) = \frac{1}{\sqrt{2 x^{2} + 1}}$ becomes $\begin{array}{l} f (x) = \frac{1}{\sqrt{x}} & \begin{array}{l} Replace the expression \\ 2 x^{2} + 1 from Step 1 with \\ x and replace H with f . \end{array} \end{array}$
Step 3 Now we have $\begin{array}{rcl} H (x) & = & f (g (x)) \\ = & (f \circ g) (x) . \end{array}$	$\begin{array}{rcll} f (g (x)) & = & f (2 x^{2} + 1) & f (x) = \frac{1}{\sqrt{x}}, g (x) = 2 x^{2} + 1 \\ = & \frac{1}{\sqrt{2 x^{2} + 1}} \\ = & H (x) \end{array}$ Note: Other choices include $g (x) = 2 x^{2}$ and $f (x) = \frac{1}{\sqrt{x + 1}}$

Practice Problem 8

Repeat Example 7 , letting $g (x) = \sqrt{2 x^{2} + 1} .$ Find f(x) and write $H (x) = (f \circ g) (x) .$

Applications of Composite Functions

5 Apply composition to practical problems.

Example 9 Calculating the Area of an Oil Spill from a Tanker

Suppose oil spills from a tanker into the Pacific Ocean and the area of the spill is a perfect circle. The radius of this oil slick increases at the rate of 2 miles per hour.

Express the area of the oil slick as a function of time.
Calculate the area covered by the oil slick in six hours.

Solution

The area A of the oil slick is a function of its radius r.

$A = f (r) = π r^{2}$

The radius is a function of time. We know that r increases at the rate of 2 miles per hour. So in t hours, the radius r of the oil slick is given by the function

$\begin{array}{l} r = g (t) = 2 t & Distance = Rate \times Time \end{array}$

Therefore, the area A of the oil slick is a function of the radius, which is itself a function of time. It is a composite function given by

$\begin{array}{rcl} A & = & f (g (t)) = f (2 t) = π {(2 t)}^{2} \\ = & 4 π t^{2} . \end{array}$
Substitute $t = 6$ into A to find the area covered by the oil in six hours.

$\begin{array}{rcl} A & = & 4 π {(6)}^{2} = 4 π (36) \\ = & 144 π \approx 452 square miles \end{array}$

Practice Problem 9

What would the result of Example 8 be if the radius of the oil slick increased at a rate of 3 miles per hour?

Example 10 Applying Composition to Sales

A car dealer offers an 8% discount off the manufacturer’s suggested retail price (MSRP) of x dollars for any new car on his lot. At the same time, the manufacturer offers a $4000 rebate for each purchase of a new car.

Write a function r(x) that represents the price after only the rebate.
Write a function d(x) that represents the price after only the dealer’s discount.
Write the functions $(r \circ d) (x)$ and $(d \circ r) (x) .$ What do they represent?
Calculate $(d \circ r) (x) - (r \circ d) (x) .$ Interpret this expression.

Solution

The MSRP is x dollars, and the manufacturer’s rebate is $4000 for each car; so

$r (x) = x - 4000$

represents the price of a car after the rebate.
The dealer’s discount is 8% of x, or 0.08x; therefore,

$d (x) = x - 0.08 x = 0.92 x$

represents the price of the car after the dealer’s discount.
1. $\begin{array}{rcll} (r \circ d) (x) & = & r (d (x)) & Apply the dealer' s discount first . \\ = & r (0.92 x) & d (x) = 0.92 x \\ = & 0.92 x - 4000 & Evaluate r (0.92 x) using r (x) = x - 4000 . \end{array}$
  
  Then $(r \circ d) (x) = 0.92 x - 4000$ represents the price of a car when the dealer’s discount is applied first and then the manufacturer’s rebate is applied.
2. $\begin{array}{rcll} (d \circ r) (x) & = & d (r (x)) & Apply the manufacturer' s rebate first . \\ = & d (x - 4000) & r (x) = x - 4000 \\ = & 0.92 (x - 4000) & Evaluate d (x - 4000) using d (x) = 0.92 x . \\ = & 0.92 - 3680 \end{array}$
  
  Thus, $(d \circ r) (x) = 0.92 x - 3680$ represents the price of a car when the manufacturer’s rebate is applied first and then the dealer’s discount is applied.
$\begin{array}{rcll} (d \circ r) (x) - (r \circ d) (x) & = & (0.92 x - 3680) - (0.92 x - 4000) & Subtract function \\ values . \\ = & $ 320 \end{array}$

This equation shows that any car, regardless of its price, will cost $320 more if you apply the rebate first and then the discount.

Practice Problem 10

Repeat Example 10 if the dealer offers a 6% discount and the manufacturer offers a $4500 rebate.

Section 2.8 Exercises

Concepts and Vocabulary

$(f \cdot g) (x) = \underline{} .$
The domain of the function $f + g$ consists of those values of x that are to the domains of f and g.
The composition of the function f with the function g is written as $f \circ g$ and is defined by $f \circ g (x) = \underline{} .$
The domain of the composite function $f \circ g$ consists of those values of x in the domain of g for which g(x) .
True or False. We always have $f \circ g = g \circ f$ .
True or False. If $f (1) = 2$ and $g (2) = 1$ , then $(f \circ g) (2) = 2$ .
True or False. The domain of $f \cdot g$ and the domain of $\frac{f}{g}$ are always the same.
True or False. A function may be decomposed into simpler functions in several different ways.

Building Skills

In Exercises 9–20, use the graphs of f and g shown in the figure to evaluate each expression.

$(f + g) (- 2)$
$(f + g) (2)$
$(f - g) (4)$
$(f - g) (- 1)$
$(f \cdot g) (- 1)$
$(f \cdot g) (2)$
$(\frac{f}{g}) (- 2)$
$(\frac{f}{g}) (2)$
$(f \circ g) (1)$
$(g \circ f) (1)$
$(f \circ g) (- 3)$
$(g \circ f) (- 3)$

In Exercises 21–24, functions f and g are given. Find each of the given values.

$(f + g) (- 1)$
$(f - g) (0)$
$(f \cdot g) (2)$
$(\frac{f}{g}) (1)$

$f (x) = 2 x; g (x) = - x$
$f (x) = 1 - x^{2}; g (x) = x + 1$
$f (x) = \frac{1}{\sqrt{x + 2}}; g (x) = 2 x + 1$
$f (x) = \frac{x}{x^{2} - 6 x + 8}; g (x) = 3 - x$

In Exercises 25–38, functions f and g are given. Find each of the following functions and state its domain.

$f + g$
$f - g$
$f \cdot g$
$\frac{f}{g}$
$\frac{g}{f}$

$f (x) = x - 3; g (x) = x^{2}$
$f (x) = 2 x - 1; g (x) = x^{2}$
$f (x) = x^{3} - 1; g (x) = 2 x^{2} + 5$
$f (x) = x^{2} - 4; g (x) = x^{2} - 6 x + 8$
$f (x) = 2 x - 1; g (x) = \sqrt{x}$
$f (x) = x - 1; g (x) = \sqrt{x}$
$f (x) = x - 6; g (x) = \sqrt{x - 3}$
$f (x) = x + 2; g (x) = \sqrt{1 - x}$
$f (x) = 1 - \frac{2}{x + 1}; g (x) = \frac{1}{x}$
$f (x) = 1 - \frac{1}{x}; g (x) = \frac{1}{x}$
$f (x) = \frac{2}{x + 1}; g (x) = \frac{x}{x + 1}$
$f (x) = \frac{5 x - 1}{x + 1}; g (x) = \frac{4 x + 10}{x + 1}$
$f (x) = \frac{x^{2}}{x + 1}; g (x) = \frac{2 x}{x^{2} - 1}$
$f (x) = \frac{x - 3}{x^{2} - 25}; g (x) = \frac{x - 3}{x^{2} + 9 x + 20}$

In Exercises 39–42, find the domain of each function.

$f \cdot g$
$\frac{f}{g}$

$f (x) = \sqrt{x - 1}; g (x) = \sqrt{5 - x}$
$f (x) = \sqrt{x - 2}; g (x) = \sqrt{x + 2}$
$f (x) = \sqrt{x + 2}; g (x) = \sqrt{9 - x^{2}}$
$f (x) = \sqrt{x^{2} - 4}; g (x) = \sqrt{25 - x^{2}}$

In Exercises 43 and 44, use each diagram to evaluate $(g \circ f) (x) .$ Then evaluate $(g \circ f) (2)$ and $(g \circ f) (- 3) .$

In Exercises 45–56, let $f (x) = 2 x + 1$ and $g (x) = 2 x^{2} - 3 .$ Evaluate each expression.

$(f \circ g) (2)$
$(g \circ f) (2)$
$(f \circ g) (- 3)$
$(g \circ f) (- 5)$
$(f \circ g) (0)$
$(g \circ f) (\frac{1}{2})$
$(f \circ g) (- c)$
$(f \circ g) (c)$
$(g \circ f) (a)$
$(g \circ f) (- a)$
$(f \circ f) (1)$
$(g \circ g) (- 1)$

In Exercises 57–66, the functions f and g are given. Find $f \circ g$ and its domain.

$f (x) = \frac{1}{x}, g (x) = 10 - 5 x$
$f (x) = \frac{1}{x}, g (x) = \sqrt{x}$
$f (x) = \sqrt{x}, g (x) = 2 x - 8$
$f (x) = \sqrt{x}, g (x) = - x$
$f (x) = \frac{2}{x + 1}; g (x) = \frac{1}{x}$
$f (x) = \frac{1}{x - 1}; g (x) = \frac{2}{x + 3}$
$f (x) = \sqrt{x - 3}; g (x) = 2 - 3 x$
$f (x) = \frac{x}{x - 1}; g (x) = 2 + 5 x$
$f (x) = | x |; g (x) = x^{2} - 1$
$f (x) = 3 x - 2; g (x) = | x - 1 |$

In Exercises 67–80, the functions f and g are given. Find each composite function and describe its domain.

$f \circ g$
$g \circ f$
$f \circ f$
$g \circ g$

$f (x) = 2 x - 3; g (x) = x + 4$
$f (x) = x - 3; g (x) = 3 x - 5$
$f (x) = 1 - 2 x; g (x) = 1 + x^{2}$
$f (x) = 2 x - 3; g (x) = 2 x^{2}$
$f (x) = 2 x^{2} + 3 x; g (x) = 2 x - 1$
$f (x) = x^{2} + 3 x; g (x) = 2 x$
$f (x) = x^{2}; g (x) = \sqrt{x}$
$f (x) = x^{2} + 2 x; g (x) = \sqrt{x + 2}$
$f (x) = \frac{1}{2 x - 1}; g (x) = \frac{1}{x^{2}}$
$f (x) = x - 1; g (x) = \frac{x}{x + 1}$
$f (x) = \sqrt{x - 1}; g (x) = \sqrt{4 - x}$
$f (x) = x^{2} - 4; g (x) = \sqrt{4 - x^{2}}$
$f (x) = \frac{1 - x}{x + 2}; g (x) = \frac{x + 3}{x - 4}$
$f (x) = \frac{x + 2}{x - 3}; g (x) = \frac{x + 1}{x - 1}$

In Exercises 81–90, express the given function H as a composition of two functions f and g such that $H (x) = (f \circ g) (x) .$

$H (x) = \sqrt{x + 2}$
$H (x) = | 3 x + 2 |$
$H (x) = {(x^{2} - 3)}^{10}$
$H (x) = \sqrt{3 x^{2}} + 5$
$H (x) = \frac{1}{3 x - 5}$
$H (x) = \frac{5}{2 x + 3}$
$H (x) = \sqrt[3]{x^{2} - 7}$
$H (x) = \sqrt[4]{x^{2} + x + 1}$
$H (x) = \frac{1}{| x^{3} - 1 |}$
$H (x) = \sqrt[3]{1 + \sqrt{x}}$

Applying the Concepts

Cost, revenue, and profit. A retailer purchases x shirts from a wholesaler at a price of $12 per shirt. Her selling price for each shirt is $22. The state has 7% sales tax. Interpret each of the following functions.
1. $f (x) = 12 x$
2. $g (x) = 22 x$
3. $h (x) = g (x) + 0.07 g (x)$
4. $P (x) = g (x) - f (x)$
Cost, revenue, and profit. The demand equation for a product is given by $x = 5000 - 5 p,$ where x is the number of units produced and sold at price p (in dollars) per unit. The cost (in dollars) of producing x units is given by $C (x) = 4 x + 12,000 .$ Express each of the following as a function of price.
1. Cost
2. Revenue
3. Profit
Cost and revenue. A manufacturer of radios estimates that his daily cost of producing x radios is given by the equation $C = 350 + 5 x .$ The equation $R = 25 x$ represents the revenue in dollars from selling x radios.
1. Write and simplify the profit function
  
  $P (x) = R (x) - C (x)$
2. Find P(20). What does the number P(20) represent?
3. How many radios should the manufacturer produce and sell to have a daily profit of $500?
4. Find the composite function $(R \circ x) (C) .$ What does this function represent?
[Hint: To find x(C), solve $C = 350 + 5 x$ for x.]
Mail order. You order merchandise worth x dollars from D-Bay Manufacturers. The company charges you sales tax of 4% of the purchase price plus shipping and handling fees of $3 plus 2% of the after-tax purchase price.
1. Write a function g(x) that represents the sales tax.
2. What does the function $h (x) = x + g (x)$ represent?
3. Write a function f(x) that represents the shipping and handling fees.
4. What does the function $T (x) = h (x) + f (x)$ represent?
Consumer issues. You work in a department store in which employees are entitled to a 30% discount on their purchases. You also have a coupon worth $5 off any item.
1. Write a function f(x) that models discounting an item by 30%.
2. Write a function g(x) that models applying the coupon.
3. Use a composition of your two functions from (a) and (b) to model your cost for an item, assuming that the clerk applies the discount first and then the coupon.
4. Use a composition of your two functions from (a) and (b) to model your cost for an item, assuming that the clerk applies the coupon first and then the discount.
5. Use the composite functions from (c) and (d) to find how much more an item costs, assuming that the clerk applies the coupon first.
Consumer issues. You work in a department store in which employees are entitled to a 20% discount on their purchases. In addition, the store is offering a 10% discount on all items.
1. Write a function f(x) that models discounting an item by 20%.
2. Write a function g(x) that models discounting an item by 10%.
3. Use a composition of your two functions from (a) and (b) to model taking the 20% discount first.
4. Use a composition of your two functions from (a) and (b) to model taking the 10% discount first.
5. Use the composite functions from (c) and (d) to decide which discount you would prefer to take first.
Test grades. Professor Harsh gave a test to his college algebra class, and nobody got more than 80 points (out of 100) on the test. One problem worth 8 points had insufficient data, so nobody could solve that problem. The professor adjusted the grades for the class by (a) increasing everyone’s score by 10% and (b) giving everyone 8 bonus points. Let x represent the original score of a student.
1. Write statements (a) and (b) as functions f(x) and g(x), respectively.
2. Find $(f \circ g) (x)$ and explain what it means.
3. Find $(g \circ f) (x)$ and explain what it means.
4. Evaluate $(f \circ g) (70)$ and $(g \circ f) (70) .$
5. Does $(f \circ g) (x) = (g \circ f) (x) ?$
6. Suppose a score of 90 or better results in an A. What is the lowest original score that will result in an A if the professor uses
  1. $(f \circ g) (x) ?$
  2. $(g \circ f) (x) ?$
Sales commissions. Henrita works as a salesperson in a department store. Her weekly salary is $200 plus a 3% bonus on weekly sales over $8000. Suppose her sales in a week are x dollars.
1. Let $f (x) = 0.03 x .$ What does this mean?
2. Let $g (x) = x - 8000 .$ What does this mean?
3. Which composite function, $(f \circ g) (x)$ or $(g \circ f) (x),$ represents Henrita’s bonus?
4. What was Henrita’s salary for the week in which her sales were $17,500?
5. What were Henrita’s sales for the week in which her salary was $521?
Enhancing a fountain. A circular fountain has a radius of x feet. A circular fence is installed around the fountain at a distance of 30 feet from its edge.
1. Write the function f(x) that represents the area of the fountain.
2. Write the function g(x) that represents the entire area enclosed by the fence.
3. What area does the function $g (x) - f (x)$ represent?
4. The cost of the fence was $4200, installed at the rate of $10.50 per running foot (per perimeter foot). You are to prepare an estimate for paving the area between the fence and the fountain at $1.75 per square foot. To the nearest dollar, what should your estimate be?
Running track design. An outdoor running track with semicircular ends and parallel sides is constructed. The length of each straight portion of the sides is 180 meters. The track has a uniform width of 4 meters throughout. The inner radius of each semicircular end is x meters.
1. Write the function f(x) that represents the area enclosed within the outer edge of the running track.
2. Write the function g(x) that represents the area of the field enclosed within the inner edge of the running track.
3. What does the function $f (x) - g (x)$ represent?
4. Suppose the inner perimeter of the track is 900 meters.
  1. Find the area of the track.
  2. Find the outer perimeter of the track.
Area of a disk. The area A of a circular disk of radius r units is given by $A = f (r) = π r^{2} .$ Suppose a metal disk is being heated and its radius r is increasing according to the equation $r = g (t) = 2 t + 1,$ where t is time in hours.
1. Find $(f \circ g) (t) .$
2. Determine A as a function of time.
3. Compare parts (a) and (b).
Volume of a balloon. The volume V of a spherical balloon of radius r inches is given by the formula $V = f (r) = \frac{4}{3} π r^{3} .$ Suppose the balloon is being inflated and its radius is increasing at the rate of 2 inches per second. That is, $r = g (t) = 2 t,$ where t is time in seconds.
1. Find $(f \circ g) (t) .$
2. Determine V as a function of time.
3. Compare parts (a) and (b).

Beyond the Basics

Let $f = {(- 2, 3), (1, 2), (2, 1), (3, 0)};$ $g = {(- 2, 0), (0, 2), (1, - 2), (3, 2)} .$ Find each function.
1. $f + g$
2. f g
3. $\frac{f}{g}$
4. $f \circ g$
Let

$\begin{array}{l} f (x) = {\begin{array}{llr} 2 x & if & - 2 \leq x \leq 1 \\ x + 1 & if & 1 < x \leq 3 \end{array}; \\ g (x) = {\begin{array}{llrll} x + 1 & if & - 3 & \leq & x \leq 2. \\ 2 x - 1 & if & 2 & \leq & x \leq 3 \end{array} \end{array}$

Find $f + g$ and its domain.
Let h(x) be any function whose domain contains $- x$ whenever it contains x. Show that
1. $f (x) = h (x) + h (- x)$ is an even function.
2. $g (x) = h (x) - h (- x)$ is an odd function.
3. h(x) can always be expressed as the sum of an even function and an odd function.
Use Exercise 105 to write each of the following functions as the sum of an even function and an odd function.
1. $h (x) = x^{2} - 2 x + 3$
2. $h (x) = 〚 x 〛 + x$
Find the domain of $f (x) = \sqrt{\frac{1 - | x |}{2 - | x |}} .$
Let $f (x) = {\begin{array}{llr} - 1 & if & - 2 \leq x \leq 0 \\ x - 1 & if & 0 < x \leq 2 \end{array} .$

Find $g (x) = f (| x |) + | f (x) |$ .

Critical Thinking / Discussion / Writing

Let $f (x) = \frac{1}{〚 x 〛}$ and $g (x) = \sqrt{x (2 - x)} .$ Find the domain of each of the following functions.
1. f(x)
2. g(x)
3. $f (x) + g (x)$
4. $\frac{f (x)}{g (x)}$
For each of the following functions, find the domain of $f \circ f .$
1. $f (x) = \frac{1}{\sqrt{- x}}$
2. $f (x) = \frac{1}{\sqrt{1 - x}}$
Even and odd functions. State whether each of the following functions is an odd function, an even function, or neither. Justify your statement.
1. The sum of two even functions
2. The sum of two odd functions
3. The sum of an even function and an odd function
4. The product of two even functions
5. The product of two odd functions
6. The product of an even function and an odd function
Even and odd functions. State whether the composite function $f \circ g$ is an odd function, an even function, or neither in the following situations.
1. f and g are odd functions.
2. f and g are even functions.
3. f is odd, and g is even.
4. f is even, and g is odd.

Getting Ready for the Next Section

1. Does the relation $R = {(- 3, 2), (- 1, 1), (1, 3), (2, 1)}$ define a function?
2. Construct a relation S by reversing the components of each ordered pair R. Does the relation S define a function?
Show that the line $y = x$ is perpendicular bisector of the line segment joining the points $P = (2, 5)$ and $P^{'} = (5, 2)$ . (This shows that the points P and $P^{'}$ are reflections of each other about the line $y = x .$ )
Solve the equation $x = 2 y + 3$ for y.
Solve the equation $x = y^{2} + 1, y \geq 0$ for y.
Solve the equation $x^{2} + y^{2} = 4, x \leq 0$ for x.
Solve the equation $2 x - \frac{1}{y} = 3$ for y.

Section 2.9 Inverse Functions

Before Starting this Section, Review

1 Vertical-line test (Section 2.4 , page 222)
2 Domain and range of a function (Section 2.4 , pages 216 and 220)
3 Composition of functions (Section 2.8 , page 284)
4 Line of symmetry (Section 2.2 , page 188)
5 Solving linear and quadratic equations (Sections 1.1 and 1.5)

Objectives

1 Define an inverse function.
2 Find the inverse function.
3 Use inverse functions to find the range of a function.
4 Apply inverse functions in the real world.

Water Pressure on Underwater Devices

The pressure at any depth in the ocean is the pressure at sea level (1 atm) plus the pressure due to the weight of the overlying water: The deeper you go, the greater the pressure. Pressure is a vital consideration in the design of all tools and vehicles that work beneath the water’s surface. Every 11 feet of depth increases the pressure by approximately 5 pounds per square inch $(1 atmosphere \approx 15 psi) .$ Computing the pressure precisely requires a somewhat complicated equation that takes into account the water’s salinity, temperature, and slight compressibility. However, a basic formula is given by “pressure equals depth times 5, divided by 11 plus 15.” Thus, we have pressure $(psi) = depth (ft) \times 5 (psi per atm) / 11 (ft per atm) + 15,$ where psi means “pounds per square inch” and atm abbreviates “atmosphere.”

For example, at 99 feet below the surface, the pressure is 60 psi. At 1 mile (5280 feet) below the surface, the pressure is $\frac{5280 \times 5}{11} + 15 = 2415 psi .$ Suppose the pressure gauge on a diving bell breaks and shows a reading of 1800 psi. We can determine the bell’s depth when the gauge failed by using an inverse function. We will return to this problem in Example 9.

Inverses

1 Define an inverse function.

Before we define inverse functions, we need to look at a special type of function called a one-to-one function.

Let f be a one-to-one function. Then the preceding definition says that for any two numbers $x_{1}$ and $x_{2}$ in the domain of f, if $x_{1} \neq x_{2},$ then $f (x_{1}) \neq f (x_{2}),$ or equivalently, if $f (x_{1}) = f (x_{2})$ then $x_{1} = x_{2} .$ That is, f is a one-to-one function if different x-values correspond to different y-values.

Figure 2.100(a) represents a one-to-one function, and Figure 2.100(b) represents a function that is not one-to-one. The relation shown in Figure 2.100(c) is not a function.

Figure 2.100

Deciding whether a diagram represents a function, a one-to-one function, or neither

Recall that with a one-to-one function, different x-values correspond to different y-values. Consequently, if a function f is not one-to-one, then there are at least two numbers $x_{1}$ and $x_{2}$ in the domain of f such that $x_{1} \neq x_{2}$ and $f (x_{1}) = f (x_{2}) .$ Geometrically, this means that the horizontal line passing through the two points $(x_{1}, f (x_{1}))$ and $(x_{2}, f (x_{2}))$ contains these two points on the graph of f. See Figure 2.101. The geometric interpretation of a one-to-one function is called the horizontal-line test.

Figure 2.101 The function `f` is *not* one-to-one

Example 1 Test for the One-to-One Property of a Function

Determine which of the following functions are one-to-one.

$f (x) = 2 x + 5$
$g (x) = x^{2} - 1$
$h (x) = 2 \sqrt{x}$

Solution

Algebraic Approach Let $x_{1}$ and $x_{2}$ be two numbers in the domain of each function.

Suppose $f (x_{1}) = f (x_{2}) .$ Then $2 x_{1} + 5 = 2 x_{2} + 5$

$\begin{array}{rcll} 2 x_{1} & = & 2 x_{2} & Subtract 5 from each side . \\ x_{1} & = & x_{2} & Divide both sides by 2 . \end{array}$

So $f (x_{1}) = f (x_{2})$ implies $x_{1} = x_{2};$ the function f is one-to-one.
Suppose $g (x_{1}) = g (x_{2}) .$ Then $x_{1}^{2} - 1 = x_{2}^{2} - 1$

$\begin{array}{rcll} x_{1}^{2} & = & x_{2}^{2} & Add 1 to both sides . \\ x_{1} & = & \pm x_{2} & Square-root property \end{array}$

This means that for two distinct numbers, $x_{1}$ and $x_{2} = - x_{1}, g (x_{1}) = g (x_{2});$ so g is not a one-to-one function.
Suppose $h (x_{1}) = h (x_{2}) .$ Then $2 \sqrt{x_{1}} = 2 \sqrt{x_{2}}$

$\begin{array}{rcll} \sqrt{x_{1}} & = & \sqrt{x_{2}} & Divide both sides by 2 . \\ x_{1} & = & x_{2} & Square both sides . \end{array}$

So $h (x_{1}) = h (x_{2})$ implies $x_{1} = x_{2};$ the function h is one-to-one.

Geometric Approach (Horizontal-line test) Figure 2.102 shows the graphs of functions $f, g,$ and h. We see that no horizontal line intersects the graphs of f (Figure 2.102(a)) or h (Figure 2.102(c)) in more than one point; therefore, the functions f and h are one-to-one. The function g is not one-to-one because the horizontal line $y = 3$ (among others) in Figure 2.102(b) intersects the graph of g at more than one point.

Practice Problem 1

Determine whether $f (x) = {(x - 1)}^{2}$ is a one-to-one function.

Side Note

The reason only a one-to-one function can have an inverse function is that if $x_{1} \neq x_{2}$ but $f (x_{1}) = y$ and $f (x_{2}) = y,$ then $f^{- 1} (y)$ must be $x_{1}$ as well as $x_{2} .$ This is not possible because $x_{1} \neq x_{2} .$

From this definition we have the following:

$Domain of f = Range of f^{- 1} and Range of f = Domain of f^{- 1}$

Warning

The notation $f^{- 1} (x)$ does not mean $\frac{1}{f (x)} .$ The expression $\frac{1}{f (x)}$ represents the reciprocal of f(x) and is sometimes written as ${[f (x)]}^{- 1} .$

Example 2 Relating the Values of a Function and Its Inverse

Assume that f is a one-to-one function.

If $f (3) = 5,$ find $f^{- 1} (5) .$
If $f^{- 1} (- 1) = 7,$ find f(7).

Solution

By definition, $f^{- 1} (y) = x$ if and only if $y = f (x) .$

Let $x = 3$ and $y = 5 .$ Now reading the definition from right to left, $5 = f (3)$ if and only if $f^{- 1} (5) = 3 .$ So $f^{- 1} (5) = 3 .$
Let $y = - 1$ and $x = 7 .$ Now $f^{- 1} (- 1) = 7$ if and only if $f (7) = - 1 .$ So $f (7) = - 1 .$

Practice Problem 2

Assume that f is a one-to-one function.
1. If $f (- 3) = 12,$ find $f^{- 1} (12) .$
2. If $f^{- 1} (4) = 9,$ find f(9).

Note the input–output diagram in the margin and Figure 2.103 for $f^{- 1} \circ f .$

For the graphical interpretation of the input–output diagram, see Figure 2.103.

Figure 2.103 suggests the following properties of inverse functions.

One interpretation of the equation $(f^{- 1} \circ f) (x) = x$ is that $f^{- 1}$ reverses the effect that f has on x. For example, let

$\begin{array}{l} f (x) = x + 2 & f adds 2 to any input x . \end{array}$

To undo what f does to x, we should subtract 2 from x. That is, the inverse of f should be

$\begin{array}{l} g (x) = x - 2 & g subtracts 2 from x . \end{array}$

Let’s verify that $g (x) = x - 2$ is indeed the inverse function of x.

$\begin{array}{rcll} (f \circ g) (x) & = & f (g (x)) & Definition of f \circ g \\ = & f (x - 2) & Replace g (x) with x - 2 . \\ = & (x - 2) + 2 & Replace x with x - 2 in f (x) = x + 2 . \\ = & x & Simplify . \end{array}$

We leave it for you to check that $(g \circ f) (x) = x .$

Example 3 Verifying Inverse Functions

Verify that the following pairs of functions are inverses of each other:

$f (x) = 2 x + 3 and g (x) = \frac{x - 3}{2}$

Solution

$\begin{array}{rcll} (f \circ g) (x) & = & f (g (x)) & Definition of f \circ g \\ = & f (\frac{x - 3}{2}) & Replace g (x) with \frac{x - 3}{2} . \\ = & 2 (\frac{x - 3}{2}) + 3 & Replace x with \frac{x - 3}{2} in f (x) = 2 x + 3 . \\ = & x & Simplify . \end{array}$

Thus, $f (g (x)) = x .$

$\begin{array}{rcll} g (x) & = & \frac{x - 3}{2} & Given function g \\ (g \circ f) (x) & = & g (f (x)) & Definition of g of f \\ = & g (2 x + 3) & Replace f (x) with 2 x + 3 . \\ = & \frac{(2 x + 3) - 3}{2} & Replace x with 2 x + 3 in g (x) = \frac{x - 3}{2} . \\ = & x & Simplify . \end{array}$

Thus, $g (f (x)) = x .$ Because $f (g (x)) = g (f (x)) = x,$ f and g are inverses of each other.

Practice Problem 3

Verify that $f (x) = 3 x - 1$ and $g (x) = \frac{x + 1}{3}$ are inverses of each other.

In Example 3, notice how the functions f and g reverse (undo) the effect of each other. The function f takes an input x, multiplies it by 2, and adds 3; g reverses (or undoes) this effect by subtracting 3 and dividing by 2. This process is illustrated in Figure 2.104. Notice that g reverses the operations performed by f and the order in which they are done.

Finding the Inverse Function

2 Find the inverse function.

Let $y = f (x)$ be a one-to-one function; then f has an inverse function. Suppose (a, b) is a point on the graph of f. Then $b = f (a) .$ This means that $a = f^{- 1} (b);$ so (b, a) is a point on the graph of $f^{- 1} .$ The points (a, b) and (b, a) are symmetric about the line $y = x,$ as shown in Figure 2.105. That is, if the graph paper is folded along the line $y = x,$ the points (a, b) and (b, a) will coincide. Therefore, we have the following property.

Figure 2.105

Relationship between points (`a`, `b`) and (`b`, `a`)

Example 4 Finding the Graph of $f^{- 1}$ from the Graph of `f`

The graph of a function f is shown in Figure 2.106. Sketch the graph of $f^{- 1} .$

Solution

By the horizontal-line test, f is a one-to-one function; so it has an inverse function.

The graph of f consists of two line segments: one joining the points $(- 3, - 5)$ and $(- 1, 2)$ and the other joining the points $(- 1, 2)$ and (4, 3).

The graph of $f^{- 1}$ is the reflection of the graph of f about the line $y = x .$ The reflections of the points $(- 3, - 5), (- 1, 2),$ and (4, 3) about the line $y = x$ are $(- 5, - 3), (2, - 1),$ and (3, 4), respectively.

The graph of $f^{- 1}$ consists of two line segments: one joining the points $(- 5, - 3)$ and $(2, - 1)$ and the other joining the points $(2, - 1)$ and (3, 4). See Figure 2.107.

Side Note

It is helpful to notice that the point (a, 0) on the x-axis, reflected about the line $y = x$ , is the point (0, a) on the y-axis, and conversely. Also, notice that points on the line $y = x$ are unaffected by reflection about the line $y = x$ .

Practice Problem 4

Use the graph of a function f in Figure 2.108 to sketch the graph of $f^{- 1} .$

Figure 2.108 Graphing $f^{- 1}$ from the graph of f

The symmetry between the graphs of f and $f^{- 1}$ about the line $y = x$ tells us that we can find an equation for the inverse function $y = f^{- 1} (x)$ from the equation of a one-to-one function $y = f (x)$ by interchanging the roles of x and y in the equation $y = f (x) .$ This results in the equation $x = f (y) .$ Then we solve the equation $x = f (y)$ for y in terms of x to get $y = f^{- 1} (x) .$

Procedure in Action

Example 5 Finding an Equation for $f^{- 1}$

OBJECTIVE	EXAMPLE
Find the inverse of a one-to-one function `f`.	Find the inverse of $f (x) = 3 x - 4 .$
Step 1 Replace `f`(`x`) with `y` in the equation defining `f`(`x`).	$\begin{array}{l} y = 3 x - 4 & Replace f (x) with y . \end{array}$
Step 2 Interchange `x` and `y`.	$\begin{array}{l} x = 3 y - 4 & Interchange x and y . \end{array}$
Step 3 Solve the equation in Step 2 for `y`.	$\begin{array}{rcll} x + 4 & = & 3 y & Add 4 to both sides . \\ \frac{x + 4}{3} & = & y & Divide both sides by 3 . \end{array}$
Step 4 Write $f^{- 1} (x)$ for `y`.	$\begin{array}{l} f^{- 1} (x) = \frac{x + 4}{3} & \begin{array}{l} We usually end with f^{- 1} (x) \\ on the left . \end{array} \end{array}$

Practice Problem 5

Find the inverse of $f (x) = - 2 x + 3 .$

Side Note

Interchanging x and y in Step 2 is done so that we can write $f^{- 1}$ in the usual format, with x as the independent variable and y as the dependent variable.

Example 6 Finding the Inverse Function

Find the inverse of the one-to-one function

$f (x) = \frac{x + 1}{x - 2}, x \neq 2 .$

Solution

Step 1 $\begin{array}{l} y = \frac{x + 1}{x - 2} & Replace f (x) with y . \end{array}$
Step 2 $\begin{array}{l} x = \frac{y + 1}{y - 2} & Interchange x and y . \end{array}$
Step 3 $\begin{array}{rcll} Solve x & = & \frac{y + 1}{y - 2} for y . & This is the most challenging step . \\ x (y - 2) & = & y + 1 & Multiply both sides by y - 2 . \\ x y - 2 x & = & y + 1 & Distributive property \\ x y - 2 x + 2 x - y & = & y + 1 + 2 x - y & Add 2 x - y to both sides . \\ x y - y & = & 2 x + 1 & Simplify . \\ y (x - 1) & = & 2 x + 1 & Factor out y . \\ y & = & \frac{2 x + 1}{x - 1} & \begin{array}{l} Divide both sides by x - 1, assuming \\ that x \neq 1 . \end{array} \end{array}$
Step 4 $\begin{array}{l} f^{- 1} (x) = \frac{2 x + 1}{x - 1}, x \neq 1 & Replace y with f^{- 1} (x) . \end{array}$

To see if our calculations are accurate, we compute $f (f^{- 1} (x))$ and $f^{- 1} (f (x)) .$

You should also check that $f (f^{- 1} (x)) = x .$

Practice Problem 6

Find the inverse of the one-to-one function

$f (x) = \frac{x}{x + 3}, x \neq - 3 .$

Finding the Range of a One-to-One Function

3 Use inverse functions to find the range of a function.

It is not always easy to determine the range of a function that is defined by an equation. However, for a one-to-one function we can find the range of f by finding the domain of $f^{- 1} .$

Example 7 Finding the Domain and Range of a One-to-One Function

Find the domain and the range of the function $f (x) = \frac{x + 1}{x - 2}$ of Example 6.

Solution

The domain of $f (x) = \frac{x + 1}{x - 2}$ is the set of all real numbers x such that $x \neq 2 .$

In interval notation, the domain of f is $(- \infty, 2) \cup (2, \infty) .$

From Example 6, $f^{- 1} (x) = \frac{2 x + 1}{x - 1}, x \neq 1;$ therefore,

$Range of f = {Domain of f}^{- 1} = {x | x \neq 1} .$

In interval notation, the range of f is $(- \infty, 1) \cup (1, \infty) .$

Practice Problem 7

Find the domain and the range of the function $f (x) = \frac{x}{x + 3} .$

If a function f is not one-to-one, then it does not have an inverse function. Sometimes by changing its domain, we can produce an interesting function that does have an inverse. (This technique is frequently used in trigonometry.) We saw in Example 1(b) that $g (x) = x^{2} - 1$ is not a one-to-one function; so g does not have an inverse function. However, the horizontal-line test shows that the function

$G (x) = x^{2} - 1, x \geq 0$

with domain $[0, \infty)$ is one-to-one. See Figure 2.109. Therefore, G has an inverse function $G^{- 1} .$

Figure 2.109 The function G has an inverse

Example 8 Finding an Inverse Function

Find the inverse of $G (x) = x^{2} - 1, x \geq 0 .$

Solution

Step 1 $\begin{array}{l} y & = & x^{2} - 1, & x \geq 0 & Replace G (x) with y . \end{array}$
Step 2 $\begin{array}{l} x & = & y^{2} - 1, & y \geq 0 & Interchange x and y . \end{array}$
Step 3 $\begin{array}{rcl} x + 1 & = & y^{2}, y \geq 0 & Add 1 to both sides . \\ y & = & \sqrt{x + 1}, because y \geq 0 & Solve for y . \end{array}$
Step 4 $\begin{array}{l} G^{- 1} (x) & = & \sqrt{x + 1} & Replace y with G^{- 1} (x) . \end{array}$

The graphs of G and $G^{- 1}$ are shown in Figure 2.110.

Figure 2.110 Graphs of `G` and $G^{- 1}$

Practice Problem 8

Find the inverse of $G (x) = x^{2} - 1, x \leq 0 .$

Applications

4 Apply inverse functions in the real world.

Functions that model real-life situations are frequently expressed as formulas with letters that remind you of the variable they represent. In finding the inverse of a function expressed by a formula, interchanging the letters could be confusing. Accordingly, we omit Step 2, keep the same letters, and just solve the formula for the other variable.

Example 9 Water Pressure on Underwater Devices

At the beginning of this section, we gave the formula for finding the pressure p (in pounds per square inch) at a depth d (in feet) below the surface of water. This formula can be written as $p = \frac{5 d}{11} + 15 .$ Suppose the pressure gauge on a diving bell breaks and shows a reading of 1800 psi. How far below the surface was the bell when the gauge failed?

Solution

We want to find the unknown depth in terms of the known pressure. This depth is given by the inverse of the function $p = \frac{5 d}{11} + 15 .$ To find the inverse, we solve the given equation for d.

$\begin{array}{rcl} p & = & \frac{5 d}{11} + 15 & Original equation \\ 11 p & = & 5 d + 165 & Multiply both sides by 11. \\ d & = & \frac{11 p}{5} - 33 & Solve for d . \end{array}$

Now we can use this formula to find the depth when the gauge reads 1800 psi. We let $p = 1800 .$

$\begin{array}{rcl} d & = & \frac{11 p}{5} - 33 & Depth from pressure equation \\ d & = & \frac{11 (1800)}{5} - 33 & Replace p with 1800. \\ d & = & 3927 \end{array}$

The device was 3927 feet below the surface when the gauge failed.

Practice Problem 9

In Example 9 , suppose the pressure gauge showed a reading of 1650 psi. Determine the depth of the bell when the gauge failed.

Section 2.9 Exercises

Concepts and Vocabulary

If no horizontal line intersects the graph of a function f in more than one point, then f is a(n) function.
A function f is one-to-one when different x-values correspond to .
If $f (x) = 3 x,$ then $f^{- 1} (x) =$ .
The graphs of a function f and its inverse $f^{- 1}$ are symmetric about the line .
True or False. If a function f has an inverse, then the domain of the inverse function is the range of f.
True or False. It is possible for a function to be its own inverse, that is, for $f = f^{- 1} .$
True or False. $f^{- 1} (x) = \frac{1}{f (x)}$
True or False. If (a, b) is a point on the graph of a one-to-one function f then (b, a) is on the graph of $f^{- 1}$ .

Building Skills

In Exercises 9–16, the graph of a function is given. Use the horizontal-line test to determine whether the function is one-to-one.

In Exercises 17–28, assume that the function f is one-to-one with domain: $(- \infty, \infty)$ .

If $f (2) = 7,$ find $f^{- 1} (7) .$
If $f^{- 1} (4) = - 7,$ find $f (- 7) .$
If $f (- 1) = 2,$ find $f^{- 1} (2) .$
If $f^{- 1} (- 3) = 5,$ find f(5).
If $f (a) = b,$ find $f^{- 1} (b) .$
If $f^{- 1} (c) = d,$ find f(d).
Find $(f^{- 1} \circ f) (337) .$
Find $(f \circ f^{- 1}) (25 π) .$
For $f (x) = 2 x - 3,$ find each of the following.
1. f(3)
2. $f^{- 1} (3)$
3. $(f \circ f^{- 1}) (19)$
4. $(f \circ f^{- 1}) (5)$
For $f (x) = x^{3},$ find each of the following.
1. f(2)
2. $f^{- 1} (8)$
3. $(f \circ f^{- 1}) (15)$
4. $(f^{- 1} \circ f) (27)$
For $f (x) = x^{3} + 1,$ find each of the following.
1. f(1)
2. $f^{- 1} (2)$
3. $(f \circ f^{- 1}) (269)$
For $g (x) = \sqrt[3]{2 x^{3} - 1},$ find each of the following.
1. g(1)
2. $g^{- 1} (1)$
3. $(g^{- 1} \circ g) (135)$

In Exercises 29–34, show that f and g are inverses of each other by verifying that $f (g (x)) = x = g (f (x)) .$

$f (x) = 3 x + 1; g (x) = \frac{x - 1}{3}$
$f (x) = 2 - 3 x; g (x) = \frac{2 - x}{3}$
$f (x) = x^{3}; g (x) = \sqrt[3]{x}$
$f (x) = \frac{1}{x}; g (x) = \frac{1}{x}$
$f (x) = \frac{x - 1}{x + 2}; g (x) = \frac{1 + 2 x}{1 - x}$
$f (x) = \frac{3 x + 2}{x - 1}; g (x) = \frac{x + 2}{x - 3}$

In Exercises 35–42, the graph of a function f is given. Sketch the graph of $f^{- 1} .$

In Exercises 43–54,

Determine algebraically whether the given function is a one-to-one function.
If the function is one-to-one, find its inverse.
Sketch the graph of the function and its inverse on the same coordinate axes.
Give the domain and intercepts of each one-to-one function and its inverse function.

$f (x) = 15 - 3 x$
$g (x) = 2 x + 5$
$f (x) = \sqrt{4 - x^{2}}$
$f (x) = - \sqrt{9 - x^{2}}$
$f (x) = \sqrt{x} + 3$
$f (x) = 4 - \sqrt{x}$
$g (x) = \sqrt[3]{x + 1}$
$h (x) = \sqrt[3]{1 - x}$
$f (x) = \frac{1}{x - 1}, x \neq 1$
$g (x) = 1 - \frac{1}{x}, x \neq 0$
$f (x) = 2 + \sqrt{x + 1}$
$f (x) = - 1 + \sqrt{x + 2}$
Find the domain and range of the function f of Exercise 33.
Find the domain and range of the function f of Exercise 34.

In Exercises 57–60, assume that the given function is one-to-one. Find the inverse of the function. Also find the domain and the range of the given function.

$f (x) = \frac{x + 1}{x - 2}, x \neq 2$
$g (x) = \frac{x + 2}{x + 1}, x \neq - 1$
$f (x) = \frac{1 - 2 x}{1 + x}, x \neq - 1$
$h (x) = \frac{x - 1}{x - 3}, x \neq 3$

In Exercises 61–68, find the inverse of each function and sketch the graph of the function and its inverse on the same coordinate axes.

$f (x) = - x^{2}, x \geq 0$
$g (x) = - x^{2}, x \leq 0$
$f (x) = | x |, x \geq 0$
$g (x) = | x |, x \leq 0$
$f (x) = x^{2} + 1, x \leq 0$
$g (x) = x^{2} + 5, x \geq 0$
$f (x) = - x^{2} + 2, x \leq 0$
$g (x) = - x^{2} - 1, x \geq 0$

Applying the Concepts

Temperature scales. Scientists use the Kelvin temperature scale, in which the lowest possible temperature (called absolute zero) is 0 K. (K denotes degrees Kelvin.)

The function $K (C) = C + 273$ gives the relationship between the Kelvin temperature (K) and Celsius temperature (C).
1. Find the inverse function of $K (C) = C + 273 .$ What does it represent?
2. Use the inverse function from (a) to find the Celsius temperature corresponding to 300 K.
3. A comfortable room temperature is $22 ° C .$ What is the corresponding Kelvin temperature?
Temperature scales. The boiling point of water is 373 K, or $212 ° F;$ the freezing point of water is 273 K, or $32 ° F .$ The relationship between Kelvin and Fahrenheit temperatures is linear.
1. Write a linear function expressing K(F) in terms of F.
2. Find the inverse of the function in part (a). What does it mean?
3. A normal human body temperature is $98.6 ° F .$ What is the corresponding Kelvin temperature?
Composition of functions. Use Exercises 69 and 70 and the composition of functions to
1. write a function that expresses F in terms of C.
2. write a function that expresses C in terms of F.
Celsius and Fahrenheit temperatures. Show that the functions in (a) and (b) of Exercise 71 are inverse functions.
Currency exchange. Alisha went to Europe last summer. She discovered that when she exchanged her U.S. dollars for euros, she received 25% fewer euros than the number of dollars she exchanged. (She got 75 euros for every 100 U.S. dollars.) When she returned to the United States, she received 25% more dollars than the number of euros she exchanged.
1. Write each conversion function.
2. Show that in part (a), the two functions are not inverse functions.
3. Does Alisha gain or lose money after converting both ways?
Hourly wages. Anwar is a short-order cook in a diner. He is paid $4 per hour plus 5% of all food sales per hour. His average hourly wage w in terms of the food sales of x dollars is $w = 4 + 0.05 x .$
1. Write the inverse function. What does it mean?
2. Use the inverse function to estimate the hourly sales at the diner if Anwar averages $12 per hour.
Hourly wages. In Exercise 74, suppose in addition that Anwar is guaranteed a minimum wage of $7 per hour.
1. Write a function expressing his hourly wage w in terms of food sales per hour. [Hint: Use a piecewise function.]
2. Does the function in part (a) have an inverse? Explain.
3. If the answer in part (b) is yes, find the inverse function. If the answer is no, restrict the domain so that the new function has an inverse.
Simple pendulum. If a pendulum is released at a certain point, the period is the time the pendulum takes to swing along its path and return to the point from which it was released. The period T (in seconds) of a simple pendulum, is a function of its length l (in feet) and is given by $T = 1.11 \sqrt{l} .$
1. Find the inverse function. What does it mean?
2. Use the inverse function to calculate the length of the pendulum, assuming that its period is two seconds.
3. The convention center in Portland, Oregon, has the longest pendulum in the United States. The pendulum’s length is 70 feet. Find the period.
Water supply. Suppose x is the height of the water above the opening at the base of a water tank. The velocity V of water that flows from the opening at the base is a function of x and is given by $V (x) = 8 \sqrt{x} .$
1. Find the inverse function. What does it mean?
2. Use the inverse function to calculate the height of the water in the tank when the flow is (i) 30 feet per second and (ii) 20 feet per second.
Physics. A projectile is fired from the origin over horizontal ground. Its altitude y (in feet) is a function of its horizontal distance x (in feet) and is given by

$y = 64 x - 2 x^{2} .$
1. Find the inverse function where the function is increasing.
2. Use the inverse function to compute the horizontal distance when the altitude of the projectile is (i) 32 feet, (ii) 256 feet, and (iii) 512 feet.
Loan repayment. Chris purchased a car at 0% interest for five years. After making a down payment, she agreed to pay the remaining $36,000 in monthly payments of $600 per month for 60 months.
1. What does the function $f (x) = 36,000 - 600 x$ represent?
2. Find the inverse of the function in part (a). What does the inverse function represent?
3. Use the inverse function to find the number of months remaining to make payments if the balance due is $22,000.
Demand function. A marketing survey finds that the number x (in millions) of computer chips the market will purchase is a function of its price p (in dollars) and is estimated by $x = 8 p^{2} - 32 p + 1200, 0 < p \leq 2 .$
1. Find the inverse function. What does it mean?
2. Use the inverse function to estimate the price of a chip if the demand is 1180.5 million chips.

Beyond the Basics

In Exercises 81 and 82, show that f and g are inverses of each other by verifying that $f (g (x)) = x = g (f (x)) .$

x 1 3 4

f(x) 3 5 2

x 3 5 2

g(x) 1 3 4

`x`	1	2	3	4
`f`(`x`)	$- 2$	0	$- 3$	1

`x`	$- 2$	0	$- 3$	1
`g`(`x`)	1	2	3	4

Let $f (x) = \sqrt{4 - x^{2}} .$
1. Sketch the graph of $y = f (x) .$
2. Is f one-to-one?
3. Find the domain and the range of f.
Let $f (x) = \frac{〚 x 〛 + 2}{〚 x 〛 - 2},$ where $〚 x 〛$ is the greatest integer function.
1. Find the domain of f.
2. Is f one-to-one?
Let $f (x) = \frac{x}{x + 1} = 1 - \frac{1}{x + 1} .$
1. Show that f is one-to-one.
2. Find the inverse of f.
3. Find the domain and the range of f.
Let $g (x) = \sqrt{1 - x^{2}}, 0 \leq x \leq 1 .$
1. Show that g is one-to-one.
2. Find the inverse of g.
3. Find the domain and the range of g.
Inverse of composition of functions. We show that if f and g have inverses, then

${(f \circ g)}^{- 1} = g^{- 1} \circ f^{- 1} . (Note the order .)$

Let $f (x) = 2 x - 1$ and $g (x) = 3 x + 4 .$
1. Find the following:
  1. $f^{- 1} (x)$
  2. $g^{- 1} (x)$
  3. $(f \circ g) (x)$
  4. $(g \circ f) (x)$
  5. ${(f \circ g)}^{- 1} (x)$
  6. ${(g \circ f)}^{- 1} (x)$
  7. $(f^{- 1} \circ g^{- 1}) (x)$
  8. $(g^{- 1} \circ f^{- 1}) (x)$
2. From part (a), conclude that
  1. ${(f \circ g)}^{- 1} = g^{- 1} \circ f^{- 1} .$
  2. ${(g \circ f)}^{- 1} = f^{- 1} \circ g^{- 1} .$
Repeat Exercise 87 for $f (x) = 2 x + 3$ and $g (x) = x^{3} - 1 .$

Critical Thinking / Discussion / Writing

Does every odd function have an inverse? Explain.
Is there an even function that has an inverse? Explain.
Does every increasing or decreasing function have an inverse? Explain.
A relation R is a set of ordered pairs (x, y). The inverse of R is the set of ordered pairs (y, x).
1. Give an example of a function whose inverse relation is not a function.
2. Give an example of a relation R whose inverse is a function.

Getting Ready for the Next Section

In Exercises 93–96, factor the given expression.

$x^{2} - x - 12$
$x^{2} - 5 x + 6$
$x^{2} + 2 x - 8$
$x^{2} + 7 x + 10$

In Exercises 97–100, solve each quadratic equation.

$x^{2} - 7 x + 12 = 0$
$x^{2} - x - 6 = 0$
$3 x^{2} + 7 x + 2 = 0$
$x^{2} - 4 x + 1 = 0$

In Exercises 101–104, use transformations on the graph $y = x^{2}$ to sketch the graph of each function.

$y = {(x + 2)}^{2} - 3$
$y = {(x - 1)}^{2} + 3$
$y = - {(x + 1)}^{2} + 2$
$y = - {(x - 3)}^{2} - 1$

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Table of Contents for Section 2.8 Combining Functions; Composite Functions

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Table of Contents for
Section 2.8 Combining Functions; Composite Functions