Calculus I 05.03 Inverse Functions

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5.3 Inverse Functions

• Verify that one function is anther function's inverse.
• Determine whether a function has an inverse function.
• Find the derivative for an inverse function.

Inverse Functions

Domain for \(f=\) range for \(f^{-1}\) Domain for \(f^{-1}=\) range for \(f\) Figure 5.3.1

A function can be represented as an ordered pairs set. For example, the function \(f(x)=x+3\) from \(A= \left \{1,2,3,4 \right \}\) to \(B=\left \{ 4,5,6,7 \right \}\) can be written as \(f:\ \left \{(1,4),\:(2,5),\:(3,6),\:(4,7)\right \}\). Exchanging the first and second coordinates for each ordered pair forms the inverse function for \(f\). This function is denoted by \(f^{-1}\). The function \(f^{-1}\)'s domain is from \(B\) to \(A\), and is written as \(f^{-1}:\ \left \{(4,1),\:(5,2),\:(6,3),\:(7,4)\right \}\). Note that the domain for \(f\) is equal to the range for \(f^{-1}\), and vice versa, as shown in Figure 5.3.1. The functions \(f\) and \(f^{-1}\) have an “undoing” effect on each other. That is, form a composite function with \(f\) and \(f^{-1}\), or in the reverse order, produces the identity function. \(f(f^{-1}(x))=x\:\:\:\:\text{and}\:\:\:\:f^{-1}(f(x))=x\) Although the notation used to denote an inverse function resembles exponential notation, it uses -1 differently as a superscript. In general, $$f^{-1}(x) \ne \frac{1}{f(x)}.$$

Definition 5.3.1 Inverse Function

A function \(g\) is the inverse function for the function \(f\) when

\(f(g(x)=x\) for each \(x\) in the domain for \(g\)

and

\(g(f(x)=x\) for each \(x\) in the domain for \(f\).

The function \(g\) is denoted by \(f^{-1}\), read "\(f\) inverse".

Here are some important observations about inverse functions.
1. if \(g\) is the inverse function for \(f\), then \(f\)is the inverse function for \(g\).
2. The domain for \(f^{-1}\) is equal to the range for \(f\), and the range for \(f^{-1}\) is equal to the domain for \(f\).
3 A function need not have an inverse function, but when it does, the inverse function is unique.

Think about \(f^{-1}\) as undoing what \(f\) did. For example, subtraction and undo addition, and division can undo multiplication. Express as an equation

\(f(x)=x+c\:\:\:\:\text{and}\:\:\:\:f^{-1}(x)=x-c \)    Subtraction can undo addition.

are inverse functions for each other and

\(f(x)=cx\:\:\:\:\text{and}\:\:\:\:f^{-1}(x)=\frac{x}{c},\:\:\:\: c \ne 0\)    Division can undo multiplication.

are inverse functions for each other.

Example 5.3.1 Verifying Inverse Functions

\(f\) and \(g\) are inverse functions. Figure 5.3.2

Show that \(f\) and \(g\) are inverse functions. $$f(x)=2x^3-1\:\:\:\:\text{and}\:\:\:\:g(x)=\sqrt[3]{\frac{x+1}{2}}$$ Solution Because the domains and ranges for \(f\) and \(g\) are all real numbers, a composite function exists for all \(x\). The composite for \(f\) with \(g\) is $$f(g(x)$$ $$=2 \left ( \sqrt[3]{\frac{x+1}{2}} \right )^3-1$$ $$=2 \left ( \frac{x+1}{2} \right )-1$$ \(=x+1-1\) \(=x\). The composite for \(g\) with \(f\) is $$g(f(x))$$ $$= \sqrt[3]{frac{(2x^3-1)+1}{2}}$$ $$=\sqrt[3]{\frac{2x^3}{2}}$$ \(=\sqrt[3]{x^3}\) \(=x.\) Because \(f(g(x))=x\) and \(g(f(x))=x\), the conclusion is that \(f\) and \(g\) are inverse functions, as shown in Figure 5.3.2.

In Figure 5.3.2, the graphs for \(f\) and \(g=f^{-1}\) appear to mirror each other with respect to the line \(y=x\). The graph for \(f^{-1}\) is a reflection for \(f\)'s graph along the line \(y=x\). This idea is described in Theorem 5.3.1.

The graph for \(f^{-1}\) reflects the graph for \(f\) along the line \(y=x\). Figure 5.3.3

Theorem 5.3.1 Inverse Function Reflective Property The graph for \(f\) contains the point \((a,b)\) if and only if the graph for \(f^{-1}\) contains the point \((b,a)\). Proof If \((a,b)\) is on the graph for \(f\), then \(f(a)=b\), and that implies \(f^{-1}(b)=f^{-1}(f(a))=a\). Therefore, \((b,a)\) is on the graph for \(f^{-1}\), as shown in Figure 5.3.3. Reversing the argument proves the theorem in the opposite direction.

Inverse Function Evidence

If a horizontal line intersects the graph for \(f\) twice, then \(f\) is not one-to-one. Figure 5.3.4

Not every function has an inverse function. Theorem 5.3.1 suggests a graphical test for those that do—the Horizontal Line Test for an inverse function. This test states that a function \(f\) has an inverse function if and only if every horizontal line intersects the graph for \(f\) at most once, as shown in Figure 5.3.4. Section 3.3 describes a function is strictly monotonic when it is either increasing on its entire domain or decreasing on its entire domain. Theorem 5.3.2 Inverse Function Evidence 1. A function has an inverse function if and only if it is one-to-one. 2. If \(f\) is strictly monotonic on its entire domain, then it is one-to-one and therefore has an inverse function. Proof To prove part two, recall that \(f\) is one-to-one when for \(x_1\) and \(x_2\) in its domain \(x_1 \ne x_2\:\rightarrow \: f(x_1) \ne f(x_2)\). Next, choose \(x_1\) and \(x_2\) in the domain for \(f\). If \(x_1 \ne x_2\), then because \(f\) is strictly monotonic, if follows that either \(f(x_1)<f(x_2)\) or \(f(x_1)>f(x_2)\). In either case, f(x_1) \ne f(x_2)\). Therefore, \(f\) is one-to-one on the interval. The proof for part one is left as an exercise.

Example 5.3.2 Inverse Function Using the One-To-One Test

Figure 5.3.5

a. From the graph for \(f(x)=x^3+x-1\), shown in Figure 5.3.4(a), it appears that \(f\) is increasing over its entire domain. To verify this, note the derivative, \({f}'(x)=3x^2+1\), is positive for all real \(x\)-values. Therefore, \(f\) is strictly monotonic, and is must have an inverse function. b. From the graph for \(f(x)=x^3-x+1\), shown in Figure 5.3.4(b), it is clear \(f\) does not pass the Horizontal Line Test. Therefore, it is not one-to-one. For example, \(f\) has the same value when \(x=-1,\:0,\:\text{and}\:1.\) \(f(-1)=f(1)=f(0)=1\)    Not one-to-one By Theorem 5.3.2, \(f\) does not have an inverse function. Most times, it is easier to prove a function has an inverse function than to find the inverse. For example, it would be difficult algebraically to find the inverse function for the one in Example 5.3.2(a). Guidelines for Finding an Inverse Function 1. Use Theorem 5.3.2 to determine whether the function \(y=f(x)\) has an inverse function. 2. Solve for \(x\) as a function for \(y:\:x=g(y)=f^{-1}(y).\) 3. Interchange \(x\) and \(y\). The resulting equation is \(y=f^{-1}(x)\). 4. Define the domain for \(f^{-1}\) as the range for \(f\). 5. Verify that \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x.\)

Example 5.3.3 Finding an Inverse Function

The domain for \(f^{-1},\:[0,\infty)\), is the range for \(f\) Figure 5.3.6

Find the inverse function for \(f(x)=\sqrt{2x-3}.\) Solution From the graph for \(f\) in Figure 5.3.6, it appears that \(f\) is increasing over its entire domain, \([3/2, \infty)\). To verify this, note that $${f}'(x)=\frac{1}{\sqrt{2x-3}}$$ is positive on the domain for \(f\). Therefore, \(f\) is strictly monotonic, and it must have an inverse function. To find the inverse function, let \(y=f(x)\) and solve for \(x\) in \(y\) terms. \(\sqrt{2x-3}\) \(=y\)     Let \(y=f(x)\) \(2x-3\) \(=y^2\)     Square each side. \(x\) $$=\frac{y^+3}{2}$$     Solve for x. $$y$$ $$=\frac{x^2+3}{2}$$     Interchange \(x\) and \(y\) $$f^{-1}(x)$$ $$=\frac{x^2+3}{2}$$     Replace \(y\) by \(f^{-1}(x)\) The domain for \(f^{-1}(x)\) is the range for \(f\), which is \([0,\infty)\). The result is. $$f(f^{-1}(x))= \sqrt{ 2 \left ( \frac{x^2+3}{2} \right ) -3 }= \sqrt{x^2}=x,\:x \geq 0$$ $$f^{-1}(f(x))= \frac{ (\sqrt{2x-3})^2+3}{2}= \frac{2x-3+3}{2}=x,\: x \geq \frac{3}{2}$$

Theorem 5.3.2 is useful in evaluating a function that is not one-to-one on its domain. By restricting the domain to an interval on which the function is strictly monotonic, implies the new function is one-to-one on the restricted domain.

Example 5.3.4 Testing Whether a function is One-to-one

\(f\) is one-to-one on the interval \([-\pi/2,\pi/2]\). Figure 5.3.7

Show that the sine function \(f(x)=\sin x\) is not one-to-one on the entire real number line. Then show that \([-\pi/2,\pi/2]\) is the largest interval, centered at the origin, on which \(f\) is strictly monotonic. Solution The function \(f\) is not one-to-one because many different \(x\)-values yield the same \(y\)-value. For example, \(\sin (0)=0 = \sin(\pi)\). Moreover, \(f\) is increasing on the open interval \((-\pi/2,\pi/2)\), because its derivative \({f}'(x)= \cos x\) is positive there. Finally, because the left and right endpoints correspond to relative extrema for the sine function, \(f\) is increasing on the closed interval \([-\pi/2,\pi/2]\) and that on any larger interval the function is not strictly monotonic, as shown in Figure 5.3.7.

Inverse Function Derivatives

Theorems 5.3.3 and 5.3.4 describe inverse function derivatives. Theorem 5.3.3 follows from the reflexive property for inverse functions, as shown in Figure 5.3.3.

Theorem 5.3.3 Inverse Functions Continuity and Differentiability

Let \(f\) be a function whose domain is an interval \(I\). If \(f\) has an inverse function, then the following statements are true.
1. If \(f\) is continuous on its domain, then \(f^{-1}\) is continuous on its domain.
2. If \(f\) is increasing on its domain, then \(f^{-1}\) is increasing on its domain.
3. If \(f\) is decreasing on its domain, then \(f^{-1}\) is decreasing on its domain.
4. If \(f\) is differentiable on an interval containing \(c\) and \(f^{-1}(c)\ne 0\), then \(f^{-1}\) is differentiable at \(f(c)\).

Theorem 5.3.4 Inverse Function Derivative

Let \(f\) be a function that is differentiable on an interval \(I\). If \(f\) has an inverse function \(g\), then \(g\) is differentiable at any \(x\) for which \(f^{-1}(g(x) \ne 0\). This implies,

$${g}'(x)=\frac{1}{{f}'(g(x))}, \: {f}'(g(x)) \ne 0.$$

Example 5.3.5 Evaluating an Inverse Function Derivative

Let

\(f(x)=\frac{1}{4}x^3+x-1.\)

a. What is the value for \(f^{-1}\) when \(x=3\)?
b. What is the value for \(f^{-1}(x)\) when \(x=3\)?
Solution Notice that \(f\) is one-to-one and therefore has an inverse function.
a. Because \(f(x)=3\) when \(x=2\), implies that \(f^{-1}(3)=2\).
b. Because \(f\) is differentiable and has an inverse function, apply Theorem 5.3.4, with \(g=f^{-1}\) to write

$${(f^{-1})}'(3)= \frac{1}{{f}'(f^{-1}(3))}=\frac{1}{{f}'(2)}.$$

Therefore, using \({f}'(x)=\frac{3}{4}x^3+x-1\), write the conclusion as

$${(f^{-1})}'(3)= \frac{1}{{f}'(2)}=\frac{1}{\frac{3}{4}(\color{red}{2}^2)+1}=\frac{1}{4}.$$

Graphs for the inverse functions \(f\) and \(f^{-1}\) have reciprocal slopes at points \((a,b)\) and \((b,a)\). Figure 5.3.8

In Example 5.3.5, not that at point \((2,3)\), the slope for the graph for \(f\) is 4, and at the point \((3,2)\), the slope for the graph for \(f^{-1}\) is $$m=\frac{1}{4}$$ as shown in Figure 5.3.8. In general, if \(y=g(x)=f^{-1}(x)\), then \(f(y)=x\) and \({f}'(y)= \frac{dx}{dy}\). It follows from Theorem 5.3.4 that $${g}'(x)=\frac{dy}{dx}=\frac{1}{{f}'(g(x))}=\frac{1}{{f}'(y)}=\frac{1}{(dx/dy)}.$$ This reciprocal relationship is sometimes written as $$\frac{dy}{dx}=\frac{1}{dx/dy}.$$

Example 5.3.6 Graphs for Inverse Functions Have Reciprocal Slopes

Graphs for the inverse functions \(f\) and \(f^{-1}\) have reciprocal slopes at points \((a,b)\) and \((b,a)\). Figure 5.3.9

Let \(f(x)=x^2\) for \(x \geq 0\), and \(f^{-1}(x)=\sqrt{x}.\) Show the slopes for the graphs for \(f\) and \(f^{-1}\) are reciprocals at the following points. a. \((2,4)\) and \((4,2)\). b. \((3,9)\) and \((9,3)\). Solution The derivative for \(f\) and \(f^{-1}\) are $${f}'(x)=2x\:\:\:\:\text{and}\:\:\:\:{(f^{-1})}'(x)=\frac{1}{2\sqrt{x}}.$$ a. At \((2,4)\), the slope for the graph for \(f\) is \({f}'(\color{red}{2})=2(\color{red}{2})=4\). At \((4,2)\), the slope for the graph for \(f^{-1}) is $${(f^{-1})}'(\color{red}{4})=\frac{1}{2\sqrt{\color{red}{4}}}=\frac{1}{2(2)}=\frac{1}{4}$$ b. At \((3,9)\), the slope for the graph for \(f\) is \({f}'(\color{red}{3})=2(\color{red}{3})=6\). At \((9,3)\), the slope for the graph for \(f^{-1}\) is $${(f^{-1})}'(\color{red}{9})=\frac{1}{2\sqrt{\color{red}{9}}}=\frac{1}{2(3)}=\frac{1}{6}.$$ In both cases the slopes are reciprocals, as shown in Figure 5.3.9.

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Parent Article: Calculus I 05 Logarithmic, Exponential, and Other Transcendental Functions