Larson and Edwards - Calculus 10/e

0. When will I need this in Calculus?

The special relationship between a function and its inverse leads to a variety of important uses in calculus. This relationship allows us to solve equations and to find points of intersection of two functions, among other applications. The concept of a one-to-one function is central to the study of inverse functions.

1. One-to-one functions

A function maps elements in one set, called the domain of the function, to elements in another set, called the range. For example, at a particular university, one can consider the function which maps maps each student (an element in the domain) to his or her college ID number (an element in the range.) The inverse of this function is a mapping in the reverse direction; it maps from ID numbers to students. In order for the inverse of a given function to be a function, the given function must have an additional property: it must be one-to-one.

A function f is one-to-one if whenever a and b are in the domain of f and $a \ne b,$ then $f(a) \ne f(b).$ This tells us that no two distinct x-values in the domain map to the same y-value. A function is not one-to-one if we can find two different x-values that are paired with the same y-value.

If we are given a graph of a set of points in a Cartesian coordinate system with horizontal axis x and vertical axis y, we can tell if this graph defines y as a function of x by the vertical line test. (See Topic 2.)

From a graph of f, we use the horizontal line test to determine if the function f is one-to-one. If f is one-to-one, a horizontal line, that is, a line with equation $y = c,$ will intersect the graph of $y = f(x)$ no more than once. If such a line intersects the graph in two or more distinct points, then those two points have the same y-value but different x-values and f is not one-to-one.

When a function is given symbolically, we can use that symbolic formulation to determine if f is one-to-one. The definition of a one-to-one function means that we can't have two distinct x-values that are paired with the same y-value. So if two y-values are the same, namely, $f(a) = f(b),$ then the x-values must also be the same; that is, $a = b.$ To check if a function given in symbols is one-to-one, we start by setting $f(a) = f(b)$ and see if that leads to $a = b.$

Example 1: Identifying one-to-one functions from a table

For each given table, assume that the complete function is given by the values in the table. Determine if the given function is one-to-one.

(a)

x	f(x)
−1	−1
0	0
1	1
2	8
3	27

(b)

x	g(x)
−3	3
−2	0
−1	0
0	−1
1	0
2	2
3	4

(c)

x	h(x)
−1.5	3
0.5	−2
10	5
11	−4
15	8
19	0

Solution:

(a) Because the list of all pairs $(x, f(x))$ are given, we can check to determine if there are two different x-values that have the same y-value. If so, then the function is not a one-to-one function. All $f(x)$ values given are different, so f is a one-to-one function. We can verify that f defines a function by observing that each x-value also appears only one time in the table.

(b) Looking at the table, we see that $g(-1) = 0$ and $g(1) = 0.$ Because $-1 \ne 1$ but $g(-1) = g(1),$ this function is not one-to-one.

(c) We observe that in all the ordered pairs of h, the second values (the y-values) are all different. This means that for any two different x-values in the domain of h, the y-values are also different, so h is one-to-one.

Example 2: Identifying one-to-one functions from a graph

For each graph, determine if the function is one-to-one. In each case, assume that the graph shows the complete set of points $(x, f(x)).$

(a)

(b)

(c)

Solution:

(a) If we look at the line $y = 0,$ we see that this line intersects the graph at more than one point. We see that the points $(-2, 0)$ and (2, 0) are both on the graph of f. This means that $f(-2) = f(2)$ so f is not one-to-one.

(b) We see from the graph that any horizontal line $y = c$ for c in the interval [1, 5] intersects the graph of f at only one point. Any other line $y = c$ does not intersect the graph. So no horizontal line intersects the graph of f in more than one point so the graph of f is one-to-one.

(c) Because no horizontal line intersects the graph of f in more than one point, the horizontal line test confirms that the graph of f is one-to-one.

Example 3: Identifying one-to-one functions

Determine if each function is one-to-one.

(a)     $f(x) = 2 - 3x$

(b)     $g(x) = \frac{1}{x + 4}$

(c)     $h(x) = 5x^2 + 1$

Solution:

(a) We start by letting a and b be values of the independent variable and setting $f(a) = f(b).$ So we set $2 - 3a = 2 - 3b.$ We subtract 2 from both sides of this equation, so $-3a = -3b.$ Dividing both sides of this equation by −3, we see that we get $a = b.$ Thus we see that if $f(a) = f(b),$ then $a = b,$ so f is one-to-one. We note that $f(x)$ is a linear function with negative slope. We can use a similar approach to show that any linear function with non-zero slope is one-to-one.

(b) We first set $g(a) = g(b);$ that is, we start with $\frac{1}{a + 4} = \frac{1}{b + 4}.$ Multiplying both sides by $(a + 4)(b + 4),$ we get $b + 4 = a + 4.$ We next subtract 4 from each side of this equation to get $b = a.$ So starting with $g(a) = g(b),$ we see that $a = b.$ This means that g is one-to-one.

(c) The graph of $h(x) = 5x^2 + 1$ is a parabola, which suggests h fails the horizontal line test and is not one-to-one. We look for two values of x, a and b with $a \ne b$ for which $h(a) = h(b).$ Let $a = -1$ and $b = 1.$ We see that $h(a) = h(-1) = 5 \cdot (-1)^2 + 1 = 6.$ But $h(b) = h(1) = 5 \cdot 1^2 + 1 = 6$ also, so h is not one-to-one.

2. Definition and properties of the inverse of a function

It is important to identify if a function f is one-to-one because if f is one-to-one, then we can define its inverse and its inverse will also be a function. Suppose the function f given by $y = f(x)$ is one-to-one. Then the function g is the inverse of f if $g(f(x)) = x$ for each x in the domain of f and $f(g(x)) = x$ for each x in the domain of g. That is, g "undoes" the function f; g takes each $f(x)$ value back to x. We denote the inverse of the function f by f⁻¹, so the definition of the inverse function provides two important relationships:

$\begin{align*} f^{-1}(f(x))& = x, \text{ for each }x \text{ in the domain of }f \\ f(f^{-1}(x))& = x, \text{ for each }x \text{ in the domain of }f^{-1}. \end{align*}$

A one-to-one function f takes an element x in the domain of f and maps it to an element y in the range of f. Then the function f⁻¹ maps y back to x. So f⁻¹ is defined for all elements in the range of f; that is, the domain of f⁻¹ is the range of f and the range of f⁻¹ is the domain of f.

Example 4: The domain and range of the inverse of a function

For each given pair of functions f and f⁻¹ give the domain and range of f and f⁻¹ and verify that $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x.$

(a) $f(x) = 4x + 3;\ f^{-1}(x) = \frac{x - 3}{4}$

(b) $f(x) = \frac{x}{x - 2};\ f^{-1}(x) = \frac{2x}{x - 1}$

Solution:

(a) Because f is a linear function, the domain of f = the range of f⁻¹ is the interval $(-\infty, \infty).$ Similarly, the range of f = the domain of f⁻¹ is also the interval $(-\infty, \infty).$ We substitute $f(x) = 4x + 3$ into the expression for f⁻¹ to see that $f^{-1}(f(x)) = f^{-1}(4x + 3) = \frac{(4x + 3) - 3}{4} = \frac{4x}{4} = x$ for all real numbers x. Similarly, $f(f^{-1}(x)) = f\left(\frac{x - 3}{4}\right) = 4\left(\frac{x - 3}{4}\right) + 3 = (x - 3) + 3 = x$ for all real numbers x.

(b) Because f is a rational function, the domain of f is all x except $x = 2,$ or, in interval notation, the interval $(-\infty, 2) \cup (2, \infty).$ Thus, the range of f⁻¹ is also the interval $(-\infty, 2) \cup (2, \infty).$ Similarly, the range of f = the domain of f⁻¹ is all x except $x = 1,$ or, in interval notation, the interval $(-\infty, 1) \cup (1, \infty).$ We see that $f^{-1}(f(x)) = f^{-1}\left(\frac{x}{x - 2}\right) = \frac{2\frac{x}{x - 2}}{\frac{x}{x - 2} - 1} = \frac{2x}{x - (x - 2)} = \frac{2x}{2} = x.$ Similarly, $f(f^{-1}(x)) = f\left(\frac{2x}{x - 1}\right) = \frac{\frac{2x}{x - 1}}{\frac{2x}{x - 1} - 2} = \frac{2x}{2x - 2(x - 1)} = \frac{2x}{2} = x.$

3. Graph of the inverse of a function

We first look at the relationship between the graphs of a function f and f⁻¹. We'll use the property that if the graph of a one-to-one function f contains a point $(x, y),$ then the graph of the inverse function contains the point $(y, x).$

Example 5: Graphing the inverse of a function

For each graph of the one-to-one function $y = f(x),$ sketch the line $y = x$ and the graph of $y = f^{-1}(x).$

(a)

(b)

Solution:

(a) The line $y = x$ is shown on the graph; it is the diagonal line through the first quadrant. To see how to obtain the graph of $y = f^{-1}(x)$ from the graph of $y = f(x),$ we can start by looking at the point (2, 0) on the graph of $y = f(x);$ this means that the point (0, 2) lies on the graph of $y = f^{-1}(x).$ The point (0, 2) is the reflection about the line $y = x$ of the point (2, 0), which is why the line $y = x$ plays an important role in visualizing an inverse function. Similarly, the point (4, 1.5) is on the graph of $y = f(x),$ so the point (1.5, 4) is on the graph of $y = f^{-1}(x).$ Reflecting the rest of the graph of f about the line $y = x$ gives us the graph shown below.

(b)

4. Finding f⁻¹ symbolically

If we have a symbolic representation of the function $y = f(x),$ we can use it to find a symbolic representation of $y = f^{-1}(x).$ To do this, we start with $y = f(x),$ interchange x and y in this equation, and then solve that equation for y.

Example 6: Finding the inverse of a function

For each function, find its inverse symbolically.

(a) $f(x) = 2 - 3x$

(b) $g(x) = \frac{1}{x + 4}$

Solution:

(a) We start with $y = 2 - 3x$ and first interchange x and y in that equation to get $x = 2 - 3y.$ We want to solve this equation for y, so we will add 3y and subtract x from both sides of this equation to get $3y = 2 - x.$ Dividing both sides by 3 gives $y = \frac{2 - x}{3}.$ So the inverse function is $f^{-1}(x) = \frac{2 - x}{3}.$

(b) Since $g(x) = \frac{1}{x + 4},$ we start with $y = \frac{1}{x + 4}$ and interchange x and y. This yields $x = \frac{1}{y + 4}.$ To solve for y, we multiply both sides by $\frac{y + 4}{x}$ to get $y + 4 = \frac{1}{x}.$ Finally, we subtract 4 from both sides of this equation to obtain the inverse function: $y = \frac{1}{x} - 4$ or $f^{-1}(x) = \frac{1}{x} - 4.$

5. Restricting the domain if f is not one-to-one

We sometimes need to find an inverse of a function that is not one-to-one. We can do this by restricting the domain of the original function so that the new function with the restricted domain is one-to-one on the restricted domain. For example, the function $f(x) = 2x^2 + 3$ is not one-to-one on its implied domain, the interval $(-\infty, \infty).$ However, if we restrict the domain to be the interval $[0, \infty)$ then the new function is one-to-one.

Because the six trigonometric functions are all periodic, none of them is one-to-one (See Topic 19.) In order to define an inverse for each of the trigonometric functions, we will restrict the domain so that the function is one-to-one on the restricted domain.

For the sine function, we want to restrict the domain to an interval over which the function passes the horizontal line test. Although we could choose other intervals, it is standard to choose the interval $\left[- \frac{\pi}{2}, \frac{\pi}{2}\right].$ Therefore, we consider the function $f(x) = \sin x, x \in \left[- \frac{\pi}{2}, \frac{\pi}{2}\right].$ This restricted sine function is one-to-one and has an inverse, called the inverse sine function and denoted $\sin^{-1} x$ or arcsin x. The domain of the inverse sine function is $[-1, 1]$ and its range is $\left[- \frac{\pi}{2}, \frac{\pi}{2}\right].$ The graphs of $y = \sin x, x \in \left[- \frac{\pi}{2}, \frac{\pi}{2}\right]$ (the solid curve) and $y = \sin^{-1} x, x \in [-1, 1]$ (the dashed curve) are shown.

For the tangent function, we again want to restrict the domain to an interval over which the function passes the horizontal line test. The standard interval is almost the same as the interval for the sine function, the interval $\left(- \frac{\pi}{2}, \frac{\pi}{2}\right),$ except the endpoints are not included because the tangent function is not defined at $\frac{\pi}{2}$ and $- \frac{\pi}{2}.$ Therefore, we consider the function $f(x) = \tan x, x \in \left(- \frac{\pi}{2}, \frac{\pi}{2}\right).$ The restricted tangent function is one-to-one and has an inverse, called the inverse tangent function and denoted $\tan^{-1} x$ or $\arctan x.$ The domain of the inverse tangent function is $(-\infty, \infty)$ and its range is $\left(- \frac{\pi}{2}, \frac{\pi}{2}\right).$ The graphs of $y = \tan x, x \in \left(- \frac{\pi}{2}, \frac{\pi}{2}\right)$ (the solid curve) and $y = \tan^{-1} x, x \in (-\infty, \infty)$ (the dashed curve) are shown.

The inverses of the other four trigonometric functions are defined in a similar way by restricting the domain of each to an appropriate interval and using the fact that, in general, the domain of f = the range of f⁻¹ and the range of f = the domain of f⁻¹.

Example 7: Finding the inverse of a function on a restricted domain

Consider the function $f(x) = 2x^2 + 3$ with domain $[0, \infty).$ Verify that f is one-to-one, find a formula for f⁻¹, and sketch the graphs of f and f⁻¹.

Solution:
We first set $f(a) = f(b)$ for $a, b \in [0, \infty).$ If $2a^2 + 3 = 2b^2 + 3,$ then $2a^2 = 2b^2$ and $a^2 = b^2.$ Because $a \geq 0$ and $b \geq 0,$ we conclude that $a = b.$ Thus f is one-to-one. To find a formula for f⁻¹, we start with $y = 2x^2 + 3, x \geq 0$ and interchange x and y to get $x = 2y^2 + 3, y \geq 0.$ We solve for y: $x - 3 = 2y^2 \text{ so } y^2 = \frac{1}{2}(x - 3).$ Because $y \geq 0,$ we solve for y by taking the positive square root: $y = \sqrt{\frac{1}{2}(x - 3)}.$ Thus $f^{-1} (x) = \sqrt{\frac{1}{2}(x - 3)}.$ The graphs of f (the solid curve) and f⁻¹ (the dashed curve) are shown on the graph below.

6. Concept Questions

1. A function f is one-to-one if whenever a and b are in the domain of f and $a \ne b,$ then $f(a) \ne f(b).$

True False

2. According to the horizontal line test, if a horizontal line intersects the graph in two or more distinct points, then the function is one-to-one.

True False

3. The function g is the inverse of f if $g(f(x)) = x$ for each x in the domain of f and $f(g(x)) = x$ for each x in the domain of g.

True False

4. All trigonometric functions are one-to-one.

True False

1	2	3	4	5	6	7	8	9	10	11	12
1/1	0/1	2/2	8/8	9/9	1/1	1/1	1/1	–/1	–/4	–/1	4/4

1	2
1/1	1/1
1/50	2/50

1	2	3	4	5	6	7	8
1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1
1/50	1/50	1/50	1/50	1/50	1/50	1/50	1/50

1	2	3	4	5	6	7	8	9
1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1	1/1
1/50	1/50	1/50	1/50	1/50	1/50	1/50	1/50	1/50

1	2	3	4
–/1	–/1	–/1	–/1
0/50	0/50	0/50	0/50

Larson and Edwards - Calculus 10/e (Homework)

Current Score : 27 / 34

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

0. When will I need this in Calculus?

1. One-to-one functions

Example 1: Identifying one-to-one functions from a table

Example 2: Identifying one-to-one functions from a graph

Example 3: Identifying one-to-one functions

2. Definition and properties of the inverse of a function

Example 4: The domain and range of the inverse of a function

3. Graph of the inverse of a function

Example 5: Graphing the inverse of a function

4. Finding f⁻¹ symbolically

Example 6: Finding the inverse of a function

5. Restricting the domain if f is not one-to-one

Example 7: Finding the inverse of a function on a restricted domain

6. Concept Questions

WebAssign

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Larson and Edwards - Calculus 10/e (Homework)

Current Score : 27 / 34

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

0. When will I need this in Calculus?

1. One-to-one functions

Example 1: Identifying one-to-one functions from a table

Example 2: Identifying one-to-one functions from a graph

Example 3: Identifying one-to-one functions

2. Definition and properties of the inverse of a function

Example 4: The domain and range of the inverse of a function

3. Graph of the inverse of a function

Example 5: Graphing the inverse of a function

4. Finding f −1 symbolically

Example 6: Finding the inverse of a function

5. Restricting the domain if f is not one-to-one

Example 7: Finding the inverse of a function on a restricted domain

6. Concept Questions

4. Finding f⁻¹ symbolically