CalcNotes Inverse Functions and Logarithms

Section 1.6 Inverse Functions and Logarithms

Subsection 1.6.1 Inverse Functions

Definition 1.6.1. One-to-one.

A function \(f\) is one-to-one (aka, injective) on a domain \(D\) if \(f(x_1)\neq f(x_2)\) whenever \(x_1\neq x_2\) are in \(D\text{.}\)

That is, a function is one-to-one if every \(y\)-value is different.

Fact 1.6.2.

A function is one-to-one if and only if every horizontal line intersects its graph at most once. If any horizontal line cuts the graph more than once, the function is not one-to-one.

Example 1.6.3.

Which of the following are one-to-one?

\(\displaystyle f(x)=x^3\)
\(\displaystyle g(x)=\sqrt{x}\)
\(\displaystyle h(x)=x^2\)
\(\displaystyle k(x)=\cos x\)

Answer.

\(f\) and \(g\) are one-to-one; \(h\) and \(k\) are not.

Definition 1.6.4. Inverse.

Let \(f\) be a one-to-one function on a domain \(D\) with range \(R\text{.}\) The inverse function \(f^{-1}\) exists and is defined by:

\begin{equation*} f^{-1}(b)=a \text{ if } f(a)=b \end{equation*}

\(f^{-1}\) has domain \(R\) and range \(D\text{.}\)

Fact 1.6.5.

If \(f\) is one-to-one with domain \(D\) and range \(R\text{,}\) then

For all \(x\) in the domain \(D\text{,}\) \((f^{-1}\circ f)(x)= x\)
For all \(y\) in the range \(R\text{,}\) \((f \circ f^{-1})(y) = y\)

Note 1.6.6.

\(f^{-1}(x) \neq \frac{1}{f(x)}\text{!!}\)

Fact 1.6.7. Finding the inverse of a function.

Given a one-to-one function \(f\text{,}\) to find a formula for \(f^{-1}(x)\text{:}\)

Start with the equation \(y=f(x)\text{.}\)
Swap \(x\)s and \(y\)s to get the equation \(x=f(y)\text{.}\)
Solve the equation \(x=f(y)\) for \(y\)
We now have \(y=f^{-1}(x)\text{.}\)

The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y=x\text{.}\)

Example 1.6.8. Find the inverse.

Let \(f(x)=\frac{x+3}{x-2}\text{.}\) Find \(f^{-1}(x)\text{.}\)

Answer.

\(f^{-1}(x) = \frac{2x+3}{x-1}\)

Fact 1.6.9. Graphing the inverse of a function.

The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y=x\text{.}\)

Example 1.6.10. Graph the inverse.

Sketch a graph of \(f(x)=\sin x, -\frac{\pi}{2}\leq x \leq \frac{\pi}{2}\) and its inverse.

A function that is not one-to-one on its natural domain can be one-to-one when restricted to a smaller domain.

Example 1.6.11. Principal fourth root.

Consider the function \(f(x)=(x-1)^4\text{,}\) \(x \geq 1\text{.}\) Sketch graphs of \(f\) and \(f^{-1}\) and find a formula for \(f^{-1}(x)\text{.}\)

Keep in mind that, for even \(n\text{,}\) \(\sqrt[n]{x^n} = |x|\text{.}\)

Solution.

We need to solve \(y=(x-1)^4\) for \(x\text{,}\) keeping in mind that the domain is \(x\geq 1\text{.}\)

\begin{align*} y \amp = (x-1)^4\\ \sqrt[4]{y} \amp = \sqrt[4]{(x-1)^4}\\ \sqrt[4]{y} \amp = |x-1|\\ \sqrt[4]{y} \amp = x-1 \amp \text{because } x\geq 1 \text{ ensures that } x-1 \text{ is always } \geq 0\\ \sqrt[4]{y} + 1 \amp = x \end{align*}

So, \(f^{-1}(x) = \sqrt[4]{x} + 1\text{,}\) with domain \(x \geq 0\text{.}\)

Example 1.6.12.

Let \(f(x)=x^2-6x\text{,}\) \(x \leq 3\text{.}\) Find a formula for \(f^{-1}(x)\text{.}\)

Solution.

We need to solve \(y=x^2-6x\) for \(x\text{,}\) under the condition that \(x \leq 3\text{.}\) Completing the square:

\begin{align*} y \amp = x^2-6x\\ y + 9 \amp = x^2-6x+9\\ y+9 \amp = (x-3)^2\\ \sqrt{y+9} \amp = |x-3|\\ \sqrt{y+9} \amp = -(x-3)\\ \sqrt{y+9} \amp = -x + 3\\ -\sqrt{y+9} + 3 \amp = x \end{align*}

So, \(f^{-1}(x) = 3 - \sqrt{x+9}\text{,}\) with domain \(x \geq -9\text{.}\)

Subsection 1.6.2 Logarithms

Definition 1.6.13. Logarithm.

The logarithm with base \(a\), for \(a\gt 0\) and \(a \neq 1\) a real number, is written \(\log_a x\) and is defined as the inverse function of the base \(a\) exponential function. That is,

\begin{equation*} y=\log_a x \hspace{3em}\text{if and only if}\hspace{3em} x=a^y \end{equation*}

The base \(e\) logarithm is called the natural logarithm and is written \(\ln x\text{,}\) i.e.,

\begin{equation*} y=\ln x \hspace{3em}\text{if and only if}\hspace{3em} x=e^y \end{equation*}

Note 1.6.14.

Compare the graphs of the exponential function and logarithm function.

Fact 1.6.15. Properties of the logarithm.

For any real numbers \(x\) and \(y\) with \(x \gt 0\) and \(y\gt 0\) and any \(r\text{:}\)

Product Rule: \(\log xy = \log x + \log y\)
Quotient Rule: \(\log \frac{x}{y} = \log x - \log y\)
Reciprocal Rule: \(\log \frac{1}{x} = -\log x\)
Power Rule: \(\log x^r = r \log x\)

(Note that these are true for any base logarithm.)

Example 1.6.16.

Express \(\displaystyle\ln\left( \frac{\sqrt[3]{7}}{50}\right)\) in terms of the natural logs of primes.

Answer.

\(\frac{1}{3}\ln 7 - 2 \ln 5 - \ln 2\)

Fact 1.6.17. Inverse properties of the logarithm and exponential.

For real numbers \(x \gt 0\) and \(a \gt 0\) with \(a\neq 1\text{,}\)

\(\displaystyle \log_a a^x = x\)
\(\displaystyle \ln e^x = x\)
\(\displaystyle a^{\log_a x} = x\)
\(\displaystyle e^{\ln x} = x\)

Example 1.6.18.

Write \(e^{2\ln x - \ln 3}\) without using \(e\) or \(\ln\text{.}\)

Example 1.6.19.

Solve for \(y\text{:}\) \(\ln (3y-5)=2t\)

Answer.

\(y = \frac{1}{3}\left(e^{2t}+5\right)\)

Example 1.6.20.

Solve for \(t\text{:}\) \(100 e^{0.02t} = 150\)

Answer.

\(t = 50 \ln 1.5\)

Fact 1.6.21. Changing base.

Observe that:

\begin{equation*} a^x = e^{\ln a^x} = e^{x\ln a} = e^{kx} \end{equation*}

where \(k=\ln a\text{,}\) so every exponential can be written with the natural base. Similarly,

\begin{equation*} \ln x = \ln (a^{\log_a x}) = (\log_a x) (\ln a) \end{equation*}

so \(\log_a x = \frac{\ln x}{\ln a}\text{.}\) This is called the change of base formula.

Example 1.6.22.

Write \(\log_5 7\) in terms of natural logs.

Answer.

\(\log_5 7 = \frac{\ln 7}{\ln 5}\)

Subsection 1.6.3 Inverse Trig Functions

By restricting the domain of the trig functions, we can make them one-to-one and thus invertible. There is a handout ¹ with images and domain restrictions.

Definition 1.6.23. Inverse Trig Functions.

\(y=\sin^{-1}x = \arcsin x\) is the value in \([-\frac{\pi}{2},\frac{\pi}{2}]\) such that \(\sin y = x\)
\(y=\cos^{-1}x = \arccos x\) is the value in \([0,\pi]\) such that \(\cos y = x\)
\(y=\tan^{-1}x = \arctan x\) is the value in \((-\frac{\pi}{2},\frac{\pi}{2})\) such that \(\tan y = x\)
\(y=\csc^{-1}x = \arccsc x\) is the value in \([-\frac{\pi}{2},0) \cup (0,\frac{\pi}{2}]\) such that \(\csc y = x\)
\(y=\sec^{-1}x = \arcsec x\) is the value in \([0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]\) such that \(\sec y = x\)
\(y=\cot^{-1}x = \arccot x\) is the value in \((0,\pi)\) such that \(\cot y = x\)

Example 1.6.24. Computing with inverse trig functions.

Find the exact values of:

\(\displaystyle \arccos \frac{\sqrt{3}}{2}\)
\(\displaystyle \sin^{-1} \left(-\frac{1}{\sqrt{2}}\right)\)
\(\displaystyle \csc^{-1} 2\)

Answer.

\(\displaystyle \frac{\pi}{6}\)
\(\displaystyle \frac{-\pi}{4}\)
\(\displaystyle \frac{\pi}{6}\)

calc1.leavesofmath.com/handouts/inverseTrigFunctions.pdf