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Section 6.3 Inverse Functions

Definition 6.3.1.

Two functions \(f\) and \(g\) are said to be inverses of each other if:

  • \(\displaystyle f(g(x)) = x \)

  • \(\displaystyle g(f(x)) = x \)

for all \(x\text{;}\) if \(g\) is the inverse of \(f\text{,}\) we often write \(f^{-1}\) instead of \(g\text{.}\)

Example 6.3.1.

How can we show that \(f(x)=x^3+7\) and \(g(x)=\sqrt[3]{x-7}\) are inverses?

Solution
\begin{align*} f(g(x)) \amp = f(\sqrt[3]{x-7})\\ \amp = \left(\sqrt[3]{x-7}\right)^3 + 7\\ \amp = x - 7 + 7 \end{align*}
\begin{align*} g(f(x)) \amp = g(x^3+7)\\ \amp = \sqrt[3]{(x^3+7) - 7}\\ \amp = \sqrt[3]{x^3}\\ \amp = x \end{align*}

Inverses are reflections of each other over the line \(y = x\text{.}\) Also, not every function has an inverse. A function has an inverse if, and only if, it is a one-to-one funcion.

One-to-One Function.

One-to-one meas that for every one input, the function only returns one output (or passes both the horizontal and vertical line tests).

Subsection 6.3.1 Obtaining Formulas for Inverse Functions

The Steps:

  1. replace \(f(x)\) with y

  2. interchange \(x\) and \(y\) variables

  3. solve for \(y\)

  4. replace \(y\) with \(f^{-1}(x)\)

Function: \(f(x) = 2x^3 + 5\)

  1. \(\displaystyle y = 2x^3 + 5\)

  2. \(\displaystyle x = 2y^3 + 5\)

  3. \(x - 5 = 2y^3\)

    \(\dfrac{x-5}{2} = y^3\)

    \(y = \sqrt[3]{\dfrac{x-5}{2}}\)

  4. \(\displaystyle f^{-1}(x) = \sqrt[3]{\dfrac{x-5}{2}}\)

Subsection 6.3.2 Derivatives of Inverse Functions

Definition 6.3.3. Formula for Derivative of an Inverse Function.
\begin{equation*} (f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))} \end{equation*}
The tangent lines of inverse functions at a given point are reciprocals of each other. So, if \(f(1) = 4\text{,}\) then \(f^{-1}(4) = 1\)

Worksheet Example Problem
A4US

Let \(f(x) = x^3 + x + 2\text{,}\) find \((f^{-1})'(4)\text{.}\)

Table 6.3.4. Inverse Function Example
Old Way New Way
\begin{align*} y \amp = x^3 + x + 2\\ x \amp = y^3 + y + 2\\ x - 2 \amp = y^3 + y\\ y^3 + y \amp = 2 - x \end{align*}
\begin{align*} f(x) \amp = x^3 + x + 2\\ f'(x) \amp = 3x^2 + 1\\ f'(1) \amp = 3(1)^2 + 1\\ \amp = 4 \end{align*}
We're stuck!
\begin{align*} (f^{-1})'(x) \amp = \dfrac{1}{f'(f^{-1}(x))} \\ (f^{-1})'(4) \amp = \dfrac{1}{f'(f^{-1}(4))} \\ \amp = \dfrac{1}{f'(1)} \\ \amp = {\frac{1}{4}} \end{align*}