MATH 144: Calculus for the Physical Sciences I

Concept 4.7.1. One-to-one functions.

A function is one-to-one (or injective) if it never takes on the same value twice; that is,

Given the graph of a function, it is easy to see whether it is one-to-one: a function is one-to-one if no horizontal line intersects its graph more than once. This is sometimes known as the horizontal line test.

Concept 4.7.2. Inverse functions.

Let $f$ be a one-to-one function with domain $A$ and range $B\text{.}$ Then its inverse function $f^{-1}$ has domain $B$ and range $A$ and is defined by

The domain of $f$ is the range of $f^{-1}\text{,}$ while the range of $f$ is the domain of $f^{-1}\text{.}$

Inverse functions satisfy:

The graph of $y=f^{-1}(x)$ can be obtained by reflecting the graph of $y=f(x)$ about the line $y=x\text{.}$

Note that $f^{-1}(x)$ is not the same as the reciprocal $\displaystyle [f(x)]^{-1} = \frac{1}{f(x)}\text{.}$

Concept 4.7.3. How to calculate $f^{-1}$ from $f$.

1. Calculate the domain and the range of $f$ (those will become the range and the domain of $f^{-1}$).
2. Check that $f$ is a one-to-one function, or restrict its domain so that it is one-to-one on this restricted domain.
3. Write $y=f(x)\text{.}$ Solve this equation for $x$ in terms of $y$ (if possible). This gives you the inverse function $x=f^{-1}(y)$ as a function of $y\text{.}$
4. To express $f^{-1}$ as a function of $x\text{,}$ interchange $x$ and $y\text{,}$ so that you get $y=f^{-1}(x)\text{.}$

Concept 4.7.4. The derivative of an inverse function.

To calculate the derivative of an inverse function $y = f^{-1}(x)\text{,}$ we use implicit differentiation. We know that:

We differentiate both sides of the relation $x = f(y)$ with respect to $x\text{,}$ treating $y$ as an unknown but differentiable function of $x\text{.}$ We get:

We solve for $y'$ to get