The Inverse Function Theorem (College Board AP® Calculus AB): Revision Note

If $f$ and $g$ are inverses, then they can be interchanged in the formula:

• $g^{\text{'}}\left(a\right)=\frac{1}{f^{\text{'}}\left(g\left(a\right)\right)}$

• $f\text{'}\left(a\right)=\frac{1}{g^{\text{'}}\left(f\left(a\right)\right)}$

Let $f$ be the function defined by $f\left(x\right)=\ln x$ for $x>0$.

Use the inverse function theorem to show that$f\text{'}\left(x\right)=\frac{1}{x}$.

Answer:

Define the inverse of $f$

Let $g\left(x\right)=f^{-1}\left(x\right)$

$g\left(x\right)=e^x$

Differentiate $g$

$g\text{'}\left(x\right)=e^x$

Use the inverse function theorem

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{g\text{'}\left(f\left(x\right)\right)}\\ & =& \frac{1}{g\text{'}\left(\ln x\right)}\\ & =& \frac{1}{e^{\ln x}}\\ & =& \frac{1}{x}\end{array}$

As $x>0$, it follows that $x\ne 0$

$f\text{'}\left(x\right)=\frac{1}{x}$

Let $f\left(x\right)={\left(3x+4\right)}^4$ and let $g$ be the inverse function of $f$.

Given that $f\left(0\right)=256$, what is the value of $g^{\text{'}}\left(256\right)$?

Answer:

Write the inverse function theorem using $f\left(x\right)$ and $f^{-1}\left(x\right)=g\left(x\right)$

$g^{\text{'}}\left(x\right)=\frac{1}{f^{\text{'}}\left(g\left(x\right)\right)}$

We are trying to find $g^{\text{'}}\left(256\right)$, so fill this in

$g^{\text{'}}\left(256\right)=\frac{1}{f^{\text{'}}\left(g\left(256\right)\right)}$

We need to find $f^{\text{'}}\left(x\right)$ and $g\left(256\right)$ to be able to calculate the value

Find $f^{\text{'}}\left(x\right)$ by differentiating $f\left(x\right)$ using the chain rule

$f^{\text{'}}\left(x\right)=4·{\left(3x+4\right)}^3·3=12{\left(3x+4\right)}^3$

Using the statement $f\left(0\right)=256$, apply the inverse to both sides to undo the function on the left-hand side

$\begin{array}{rcl}f\left(0\right)& =& 256\\ f^{-1}\left(256\right)& =& 0\end{array}$

$g\left(x\right)$ is the inverse of $f\left(x\right)$

$g\left(256\right)=0$

We now have all the information we need, so substitute these in

$g^{\text{'}}\left(256\right)=\frac{1}{f^{\text{'}}\left(g\left(256\right)\right)}=\frac{1}{f^{\text{'}}\left(0\right)}=\frac{1}{12{\left(3\left(0\right)+4\right)}^3}=\frac{1}{12{\left(4\right)}^3}=\frac{1}{768}$

$g^{\text{'}}\left(256\right)=\frac{1}{768}$

The function $f$ is defined by $f\left(x\right)=x^3+2x-10$.

Show that the inverse of $f\left(x\right)$ exists, and then find the derivative of the inverse of $f\left(x\right)$ at the point where $x=125$.

Answer:

First check that the inverse exists

$f^{\text{'}}\left(x\right)=3x^2+2$, which is always positive

So $f\left(x\right)$ is always strictly increasing, which means it is a one-to-one function

Therefore $f^{-1}\left(x\right)$ exists

Use the inverse function theorem

${\left(f^{-1}\right)}^{\text{'}}\left(a\right)=\frac{1}{f^{\text{'}}\left(f^{-1}\left(a\right)\right)}$

We are trying to find the derivative of the inverse at $x=125$, or ${\left(f^{-1}\right)}^{\text{'}}\left(125\right)$

Fill this in

${\left(f^{-1}\right)}^{\text{'}}\left(125\right)=\frac{1}{f^{\text{'}}\left(f^{-1}\left(125\right)\right)}$

We already know from the first step that $f^{\text{'}}\left(x\right)=3x^2+2$

So we just need to find $f^{-1}\left(125\right)$; this is the inverse of $f$, when $x=125$

To find an equation for the inverse of $f$, simply switch $x$ and $y$

$f\left(x\right)=y=x^3+2x-10$

Inverse: $x=y^3+2y-10$

Find the inverse when $x=125$

$\begin{array}{rcl}125& =& y^3+2y-10\\ 0& =& y^3+2y-135\end{array}$

Solve the cubic equation

• If it were a calculator question, you could solve the cubic equation on your calculator

• If it were a non-calculator question, you could test values for $y$ using factors of 135

$y=5$

Fill this in

${\left(f^{-1}\right)}^{\text{'}}\left(125\right)=\frac{1}{f^{\text{'}}\left(f^{-1}\left(125\right)\right)}=\frac{1}{f^{\text{'}}\left(5\right)}=\frac{1}{3{\left(5\right)}^2+2}=\frac{1}{77}$

${\left(f^{-1}\right)}^{\text{'}}\left(125\right)=\frac{1}{77}$