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Section 4.7 Inverse functions

In this section we review the concept of inverse functions, and use implicit differentiation to calculate the derivative of an inverse function. This will helpful to study exponentials, logarithms, and inverse trigonometric functions in the next few sections.

Subsection 4.7.1 Instructional video

Subsection 4.7.2 Key concepts

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,

\begin{equation*} f(x_1) \neq f(x_2) \qquad \text{whenever $x_1 \neq x_2$.} \end{equation*}

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

\begin{equation*} y = f(x) \qquad \text{if and only if} \qquad f^{-1}(y) = x. \end{equation*}

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:

\begin{align*} f^{-1}(f(x)) =\amp x \qquad \text{for $x$ in the domain of $f$,}\\ f(f^{-1}(x)) =\amp x \qquad \text{for $x$ in the domain of $f^{-1}$}. \end{align*}

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:

\begin{equation*} y = f^{-1}(x) \qquad \text{if and only if} \qquad x = f(y). \end{equation*}

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:

\begin{equation*} 1 = f'(y) \cdot \frac{dy}{dx}. \end{equation*}

We solve for \(y'\) to get

\begin{equation*} y' = \frac{1}{f'(y)} = \frac{1}{f' (f^{-1}(x) )}. \end{equation*}

Further readings 4.7.3 Further readings