Inverse functions

Section 1.10 Inverse functions

Definition 1.10.1. One-to-one function.

A function \(f\) is one-to-one on its domain \(D\) if \(f(x_1)\ne f(x_2)\) for all \(x_1, x_2\in D\) with \(x_1\ne x_2\text{.}\) We express this with logical notation as

\begin{equation*} x_1\ne x_2\implies f(x_1)\ne f(x_2)\text{,} \end{equation*}

or equivalently, using the contrapositive,

\begin{equation*} f(x_1)=f(x_2)\implies x_1=x_2\text{.} \end{equation*}

Definition 1.10.2. Inverse function.

Suppose \(f\) is one-to-one on its domain \(D\text{,}\) and let \(R\) be the range of \(f\text{.}\) The inverse function of \(f\) is the function \(f^{-1}\) with domain \(R\) defined by the following rule: for any \(b\in R\text{,}\) we define \(f^{-1}(b)\) to be the unique \(a\in D\) such that \(f(a)=b\text{.}\) In other words, \(f^{-1}\) is the unique function with domain \(R\) satisfying

\begin{equation} f^{-1}(b)=a \iff b=f(a)\tag{1.10.1} \end{equation}

for all elements \(b\) in \(R\text{.}\)

Theorem 1.10.3. Horizontal line test.

Let \(f\) be a function with domain \(D\text{,}\) and let \(\mathcal{C}\) be the graph of \(f\text{.}\) The function \(f\) is one-to-one on \(D\) if and only if for all \(c\in\R\) the horizontal line \(y=c\) intersects \(\mathcal{C}\) in at most one point.

Definition 1.10.4. Monotonic.

Let \(f\) be a a real-valued function with domain \(D\text{.}\)

The function \(f\) is increasing on \(D\) if \(f(x_1)\lt f(x_2)\) for all \(x_1, x_2\in D\) with \(x_1\lt x_2\text{.}\) Using logical notation:

\begin{equation*} x_1\lt x_2\implies f(x_1)\lt f(x_2)\text{.} \end{equation*}
The function \(f\) is decreasing on \(D\) if \(f(x_1)>f(x_2)\) for all \(x_1, x_2\in D\) with \(x_1\lt x_2\text{.}\) Using logical notation:

\begin{equation*} x_1\lt x_2\implies f(x_1)>f(x_2)\text{.} \end{equation*}
The function \(f\) is monotonic on \(D\) if \(f\) is increasing on \(X\) or \(f\) is decreasing on \(X\text{.}\)

Theorem 1.10.5. Monotonic functions are one-to-one.

If \(f\) is monotonic on its domain \(D\text{,}\) then \(f\) is one-to-one on \(D\) and hence has an inverse function.

Example 1.10.6. Computing inverse.

Let \(f(x)=x^2+1\text{.}\)

Show that \(f\) is not one-to-one on \(D=(-\infty, \infty)\text{.}\)
Show that \(f\) is one-to-one on \(D=(-\infty, 0]\) and compute an inverse of \(f\) with respect to this domain.

Solution.

Using simple properties of quadratic functions, we know that \(f\) is decreasing on \((-\infty, 0]\) and increasing on \([0,\infty)\text{.}\) It follows that \(f\) is one-to-one on either of those smaller subintervals. Since \(f(-2)=f(2)=5\text{,}\) and more generally \(f(-x)=f(x)\) for all \(x\in \R \text{,}\) we see that \(f\) is not one-to-one on \(\R=(-\infty,\infty)\text{.}\)

Now restrict to the domain \(D=(-\infty, 0]\text{.}\) By definition, the domain of \(f^{-1}\) is the range of \(y\text{,}\) which is easily seen to be \(R=[1,\infty)\text{.}\) Given \(y\in [1,\infty)\text{,}\) to compute \(f^{-1}(y)\) we must find the unique \(x\in D=(-\infty,0]\) satisfying \(f(x)=y\text{.}\) Solving this equation for \(x\text{,}\) we see that

\begin{align*} f(x)=y \amp \iff x^2+1=y\\ \amp \iff x^2=y-1 \\ \amp \iff x=\pm \sqrt{y-1}\text{.} \end{align*}

Since \(\sqrt{y-1}\) is always nonnegative, and since we seek \(x\) lying in \(D=(-\infty,0]\text{,}\) we see that we must have \(x=-\sqrt{y-1}\text{,}\) and thus

\begin{equation*} f^{-1}(y)=-\sqrt{y-1}\text{.} \end{equation*}

Theorem 1.10.7. Inverse function compendium.

Let \(f\) be one-to-one on its domain \(D\text{,}\) let \(R\) be the range of \(f\text{,}\) and let \(f^{-1}\) be the inverse of \(f\text{.}\)

\(f^{-1}(b)=a \iff b=f(a)\) for all \(b\in R\text{.}\)
The domain of \(f^{-1}\) is \(R\text{,}\) the range of \(f\text{;}\) the range of \(f^{-1}\) is \(D\text{,}\) the domain of \(f\text{.}\)
We have

\begin{align} f^{-1}(f(a))\amp =a \text{ for all } a\in X\tag{1.10.2}\\ f(f^{-1}(b))\amp =b \text{ for all } b\in Y\text{.}\tag{1.10.3} \end{align}
The point \(P=(x,y)\) is on the graph of \(f\) if and only if the point \(Q=(y,x)\) is on the graph of \(f^{-1}\text{.}\)
The graph of \(f^{-1}\) is the reflection of the graph of \(f\) through the line \(y=x\text{.}\)

Theorem 1.10.8. Derivative formula for inverses.

Assume \(f\) is one-to-one and differentiable on the interval \(I\text{,}\) and that \(f'(x)\ne 0\) for all \(x\in I\text{.}\) Let \(J\) be the range of \(f\text{.}\)

The inverse function \(f^{-1}\) is differentiable on \(J\text{.}\)
We have

\begin{equation} (f^{-1})'(b)=\frac{1}{f'(f^{-1}(b))}\tag{1.10.4} \end{equation}

for all \(b\in J\text{.}\)

Proof.

We will not prove (1), which is the more difficult part of this theorem. Assuming \((1)\) is true, we prove (2) using the chain rule as follows. By (2) of Theorem 1.10.7, we have \(f^{-1}(f(x))=x\) for all \(x\) in the domain of \(f\text{.}\) Taking derivatives of both sides of this equation, we see that

\begin{align*} \frac{d}{dx} (f^{-1}(f(x)))=\frac{d}{dx} x\amp \implies (f^{-1})'(f(x))f'(x)=1 \amp \text{(chain rule)}\\ \amp \implies f^{-1}(f(x))=\frac{1}{f'(x)} \amp (f'(x)\ne 0)\text{.} \end{align*}

Equivalently, since for any \(b\in J\) we have \(f^{-1}(b)=a\) for some unique \(a\in I\) satisfying \(f(a)=b\text{,}\) we see that

\begin{align*} (f^{-1})'(b) \amp = (f^{-1})'(f(a)) \\ \amp = \frac{1}{f'(a)} \amp \text{(by above)}\\ \amp = \frac{1}{f'(f^{-1}(b))} \amp (a=f^{-1}(b))\text{.} \end{align*}

Procedure 1.10.9. Computing \((f^{-1})'(b)\).

Let \(f\) be one-to-one on its domain \(D\text{,}\) let \(R\) be the range of \(f\text{,}\) and let \(f^{-1}\) be the inverse function of \(f\text{.}\) Given \(b\in R\text{,}\) to compute \((f^{-1})'(b)\text{,}\) proceed as follows.

Compute \(f^{-1}(b)=a\text{.}\) This is often done by inspection: i.e., by finding the unique \(a\) such that \(f(a)=b\text{.}\)
Compute \(f'(x)\text{.}\)
Use (1.10.4) to conclude \((f^{-1})'(b)=1/f'(f^{-1}(b))=1/f'(a)\text{.}\)

Example 1.10.10. Derivative of inverse.

Let \(f\) be defined as \(f(x)=x^5+x^3+3x-5\) on the domain \(D=\R\) of all real numbers.

Show that \(f\) has an inverse function.
Plot three points on the graph of \(f^{-1}\text{.}\)
Compute \((f^{-1})'(-5)\) and \((f^{-1})'(-10)\text{.}\)

Solution.

Since \(f'(x)=5x^4+3x^2+3\) is clearly positive for all \(x\in \R\text{,}\) we see that \(f\) is increasing on its domain, and hence is one-to-one. Thus \(f\) has an inverse function.
Short of actually plotting points, we can at least produce a table of values for \(f^{-1}\) using a table of values of \(f\text{,}\) using (4) of Theorem 1.10.7.

\begin{align*} \begin{array}{c|c} x\amp f(x) \\ \hline -1\amp -10\\ 0 \amp -5 \\ 1 \amp -2 \end{array} \amp\implies \begin{array}{c|c} x\amp f^{-1}(x) \\ \hline -10\amp -1\\ -5 \amp 0 \\ -2\amp 1 \end{array} \end{align*}
For general \(b\) in the domain of \(f^{-1}\) (or equivalently, range of \(f\)), we have by Theorem 1.10.8

\begin{align*} f^{-1}(b) \amp = \frac{1}{f'(f^{-1}(b))}\\ \amp = \frac{1}{5(f^{-1}(b))^4+3(f^{-1}(b))^2+3} \amp (f'(x)=5x^4+3x^2+3)\text{.} \end{align*}

Thus to compute \((f^{-1})'(-5)\) and \((f^{-1})'(-10)\text{,}\) it remains only to compute \(f^{-1}(-5)\) and \(f^{-1}(-10)\) and plug these into the last expression above. We see using our table of values in (b) that \(f^{-1}(-5)=0\) and \(f^{-1}(-10)=-1\text{.}\) Thus

\begin{align*} (f^{-1})'(-5) \amp = \frac{1}{(5\cdot 0+3\cdot 0+3)}=\frac{1}{3} \\ (f^{-1})'(-10) \amp = \frac{1}{(5\cdot (-1)^4+3\cdot (-1)^2+3)}=\frac{1}{11} \text{.} \end{align*}

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