Consider the function $f(x) = x^2$ for $x \ge 0.$ Find $\left[f^{-1}\right]'(4)$.

Solution.
We’ll solve this using three approaches to illustrate various ways you can find the result.

Method 1. Use the explicit inverse function $g(x) = f^{-1}(x) = \sqrt{x}.$
This is the method we would have used before we developed the results on this screen. Since $f(x) = x^2,$ we know immediately that its inverse function is $g(x) = f^{-1}(x) = \sqrt{x}.$ We further know that $g'(x) = \dfrac{1}{2\sqrt{x}}$, and so
\[\left[f^{-1}\right]'(4) = g'(4) = \dfrac{1}{2 \sqrt{4}} = \dfrac{1}{4} \quad \cmark \]

Let’s see how the Inverse Function theorem leads to the same result.

Method 2. Use the Inverse Function theorem.
\[\big[f^{-1}\big]'(4) = \frac{1}{f’\Big(f^{-1}(4)\Big)}\] The input to $f’$ in the denominator is $f^{-1}(4),$ so let’s figure that out first: What input value of x (where $x \ge 0$) gives an output value of $f(x) = x^2 = 4$? We can answer “by inspection”: we look at the equation, and simply immediately know that $x = 2$: $f(2) = 4.$ Hence
\[f^{-1}(4) = 2 \quad \blacktriangleleft\]

Next, we find the derivative of f:
\begin{align*}
f(x) &= x^2 \\[8px] f'(x) &= 2x \quad \blacktriangleleft
\end{align*}
Then finally
\begin{align*}
\big[f^{-1}\big]'(4) &= \frac{1}{f’\Big(f^{-1}(4)\Big)} \\[8px] &= \frac{1}{f'(2)} \\[8px] &= \frac{1}{2(2)} \quad [\text{since }f'(x) = 2x] \\[8px] &= \frac{1}{4} \quad \cmark
\end{align*}
Method 3. Start with the defining relation for inverse functions, and use the Chain Rule.
Let’s say you didn’t immediately remember the Inverse Function theorem. (It happens!) If you’re given an expression for the function f, you can always proceed from the defining relation for inverse functions, $f^{-1}\big(f(x)\big) = x.$ We’ll use the notation $g(x) = f^{-1}(x)$ since it’s a little easier to write and track everything without the –1’s floating around:
\begin{align*}
f^{-1}\left(x^2 \right) = g\left(x^2\right) &= x \\[8px] \Big[g\left(x^2\right)\Big]’ &= [x]’ \\[8px] g’\left(x^2\right) \cdot (2x) &= 1 \quad [\text{[Remember the Chain Rule!]}] \\[8px] g’\left(x^2\right) &= \frac{1}{2x}
\end{align*}
Notice that we actually just developed the Inverse Function theorem for the particular function $f(x) = x^2,$ which is what always happens with this method.
We still have to realize that since we have $g’\left(x^2\right)$ on the left-hand side of the equation and we want $g'(4),$ that means we have $x=2$:
\begin{align*}
g’\left(x^2\right) &= \frac{1}{2x} \\[8px] g'(4) &= \frac{1}{2(2)} \\[8px] &= \frac{1}{4} \quad \cmark
\end{align*}

Let $p(x) = x^3 + 3x + 1$. Use the Inverse Function theorem to find $[p^{-1}]'(5)$
Hint: Use inspection to determine $p^{-1}(5)$: What input value of x gives $p(x) = x^3 + 3x + 1 = 5$? Good choices to start guessing are often $x= 0 $ and $x = 1.$

Solution.
The Inverse Function theorem says
\[\left[p^{-1}\right]'(5) = \frac{1}{p’\big(p^{-1}(5)\big)}\] To find $p^{-1}(5)$, we can do a little bit of guess work. Remember that we’re looking for x such that $x^3 + 3x + 1 = 5$.

Let’s try $x=0$:
\[p(0) = 0 + 0 + 1 = 1 \quad \xmark\] That’s not what we were looking for. Let’s next try $x=1$:
\[p(1) = 1 + 3 + 1 = 5 \quad \blacktriangleleft\] Ah, that’s it: $p(1) = 5.$ Therefore, $p^{-1}(5) = 1$.

Next, let’s find $p'(x)$:
\begin{align*}
p(x) &= x^3 + 3x + 1 \\[8px] p'(x) &= 3x^2 + 3
\end{align*}
Then
\[p'(1) = 3(1)^2 + 3 = 6 \quad \blacktriangleleft\] Now returning to the Inverse Function theorem:
\begin{align*}
\left[p^{-1}\right]'(5) &= \frac{1}{p'(p^{-1}(5))} \\[8px] &= \frac{1}{p'(1)} \\[8px] &= \frac{1}{6} \quad \cmark
\end{align*}