section 6.1
Table of Contents
Inverse Functions
If you’re enrolled in Math/CS 321, see this footnote^{1}
Recall from algebra/precalc/trig: #
 A function $f$ is called onetoone if it never takes the same value twice, that ifs $f(a) \ne f(b)$ if $a \ne b$.
 The Horizontal Line Test A function is onetoone if and only if no horizontal line intersects a graph more than once.
Inverse Functions #
Let $f$ be a onetoone function with domain $A$ and range $B$, then its inverse function, $f^{1}$ has domain $B$ and range $A$ and is defined by $$ f^{1}(y) = x \iff f(x) = y $$ for any $y$ in $B$.
 the domain of $f$ is the range of $f^{1}$
 the range of $f$ is the domain of $f^{1}$
that is, the $x$’s become the $y$’s and vice versa.
Finding the inverse of a function: the process: #
 Write $y=f(x)$
 Switch the roles of $x$ and $y$ (swap the variables)
 Solve this new equation for the new $y$.
 Write $f^{1}$ as this function, $y=f^{1}(x)$
Example #
Find the inverse of $f(x) = x^3  5$, then graph both $f$ and $f^{1}$ on the same axes.
Answer
Click to reveal the answer.
Do some algebra and confirm $f^{1} = \sqrt[3]{x+5}$Example #
Find the inverse of $f(x) = \sqrt{x  1}$, then graph both $f$ and $f^{1}$ on the same axes.
Answer
Click to reveal the answer.
Do some algebra and confirm $f^{1} = x^2 + 1$. **Note** Because the range of $f$ is $[0, \infty)$, the domain of $f^{1}$ is only $[0,\infty)$, meaning it's only the righthalf of the parabola! (otherwise it's not onetoone).Takeaway: Inverse function graphs are symmetric over the line $y=x$.
The derivative of inverse functions #
If $f$ is a onetoone function with inverse function $f^{1}$, and $f’(f^{1}(a))\ne 0$, then the inverse function is differentiable at $a$ and: $$ (f^{1})’(a)=\frac{1}{f’(f^{1}(a))} $$
Example #
Consider $f(x) = x^2$ on the restricted domain $[0, \infty)$.
 Show that $f$ has an inverse and that domain.
 Find $(f^{1})’(1)$ using the above formula.
 Find $(f^{1})’(4)$ using the above formula.
 Find the inverse of $f$ and show that the computed derivatives match what we expect.

Note that in 321 we define a function by a domain and a codomain. In that context, $f: A \to B$ is invertible if and only if $f$ is both onetoone (injective) and $f$ is onto (surjective). In calculus we’re just considering the case where $B=f(A)$ and so the functions that we are considering are by their very nature already surjective. That’s why we only need to test injectivity. ↩︎