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Section 5.1 Section 5.1 Inverse Functions

Subsection 1. Use new vocabulary

Subsubsection Definitions

Write a definition or description for each term:

  1. Inverse function
  2. Inverse function notation
  3. Horizontal line test
  4. One-to-one function
  5. Symmetric about \(y=x\)

Subsubsection Exercise

Illustrate each term above for the following situation:

The sales tax \(T\) on an item that costs \(p\) dollars is given by the function \(T=f(p)=1.15p\)

Answer
  1. The inverse function gives the price \(p\) of an item whose sales tax is \(T\) dollars.
  2. \(\displaystyle p=f^{-1}(T)\)
  3. The graph of \(T=f(p)=1.15p\) is linear, and so passes the horizontal line test.
  4. A function that passes the horizontal line test is one-to-one: for each value of \(p\) there is only one value of \(T\text{,}\) and vice versa.
  5. lines

Subsection 2. Solve an equation for a variable

When finding a formual for an inverse function, we need to solve for one variable in terms of the other.

Subsubsection Examples

Example 5.2.

Solve \(~y=\sqrt{x^3-4}~\) for \(x\) in terms of \(y\text{.}\)

Solution
\begin{align*} y \amp =\sqrt{x^3-4} \amp \amp \blert{\text{Square both sides.}}\\ y^2 \amp = x^3-4 \amp \amp \blert{\text{Add 4 to both sides.}}\\ y^2 + 4 \amp = x^3 \amp \amp \blert{\text{Take cube roots.}}\\ \sqrt[3]{y^2+4} \amp = x \end{align*}
Example 5.3.

Solve \(~y=\dfrac{x+1}{x-2}~\) for \(x\) in terms of \(y\text{.}\)

Solution
\begin{align*} y \amp =\dfrac{x+1}{x-2} \amp \amp \blert{\text{Clear the fraction.}}\\ y(x-2) \amp = x+1 \amp \amp \blert{\text{Apply the distributive law.}}\\ xy-2y \amp = x+1 \amp \amp \blert{\text{Collect}~x~\text{terms.}}\\ xy-x \amp = 1+2y \amp \amp \blert{\text{Factor out}~x.}\\ x(y-1) \amp = 1+2y \amp \amp \blert{\text{Divide by}~y-1.}\\ x \amp = \dfrac{1+2y}{y-1} \end{align*}

Subsubsection Exercises

Solve \(~y=5-4\sqrt{x+2}~\) for \(x\) in terms of \(y\text{.}\)

Answer
\(x=\dfrac{(y-5)^2}{16}-2\)

Solve \(~y=\dfrac{3}{\sqrt[3]{x+6}}~\) for \(x\) in terms of \(y\text{.}\)

Answer
\(x=\dfrac{27}{y^3}-6\)

Solve \(~y=(2x-3)^3+1~\) for \(x\) in terms of \(y\text{.}\)

Answer
\(x=\dfrac{1}{2}\left(3+\sqrt[3]{y-1}\right)\)

Solve \(~y=\dfrac{3x-1}{2x+1}~\) for \(x\) in terms of \(y\text{.}\)

Answer
\(x=\dfrac{y+1}{3-2y}\)

Subsection 3. Use function notation

Keep in mind that the notation for an inverse function, \(f^{-1}(x)\text{,}\) does not mean the reciprocal of the function. This is a different use of the notation from how it is used as an exponent.

Subsubsection Examples

Example 5.8.

\(f(x)=2x-3~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Solution
  1. \(g(x)\) is the reciprocal of \(f(x)\text{:}\) \(g(x)=\dfrac{1}{2x-3}\)
  2. \(h(x)\) is the negative or opposite of \(f(x)\text{:}\) \(h(x)=3-2x\)
  3. To find the inverse function for \(f(x)\text{,}\) we write \(~y = 2x-3~\) and solve for \(x\text{.}\)
    \begin{align*} y \amp = 2x-3 \amp \amp \blert{\text{Add 3 to both sides.}}\\ y+3 \amp = 2x \amp \amp \blert{\text{Divide both sides by 2.}}\\ \dfrac{y+3}{2} \amp = x \end{align*}
    We can use any variables for a function, so we switch back to \(x\) for the input and \(y\) for the output: \(~y = \dfrac{x+3}{2}~\text{.}\) Thus, the inverse function is
    \begin{equation*} j(x)=\dfrac{x+3}{2} \end{equation*}
Example 5.9.

\(f(x)=\sqrt[3]{x-4}~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Solution
  1. \(g(x)\) is the reciprocal of \(f(x)\text{:}\) \(~g(x)=\dfrac{1}{\sqrt[3]{x-4}}\)
  2. \(h(x)\) is the negative or opposite of \(f(x)\text{:}\) \(~h(x)=-\sqrt[3]{x-4}\)
  3. To find the inverse function for \(f(x)\text{,}\) we write \(~y = \sqrt[3]{x-4}~\) and solve for \(x\text{.}\)
    \begin{align*} y \amp = \sqrt[3]{x-4} \amp \amp \blert{\text{Cube both sides.}}\\ y^3 \amp = x-4 \amp \amp \blert{\text{Add 4 to both sides.}}\\ y^3+4 \amp = x \end{align*}
    We can use any variables for a function, so we switch back to \(x\) for the input and \(y\) for the output:
    \begin{equation*} j(x)=x^3+4 \end{equation*}

Subsubsection Exercise

\(f(x)=2-\dfrac{1}{2}x~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Answer
  1. \(\displaystyle g(x)=\dfrac{2}{4-x}\)
  2. \(\displaystyle h(x)=\dfrac{1}{2}x-2\)
  3. \(\displaystyle j(x)=4-2x\)

\(f(x)=\dfrac{1}{x+2}~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Answer
  1. \(\displaystyle g(x)=x+2\)
  2. \(\displaystyle h(x)=\dfrac{-1}{x+2}\)
  3. \(\displaystyle j(x)=\dfrac{1-2x}{x}\)

\(f(x)=x^{3/4}~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Answer
  1. \(\displaystyle g(x)=x^{-3/4}\)
  2. \(\displaystyle h(x)=-x^{3/4}\)
  3. \(\displaystyle j(x)=x^{4/3}\)

\(f(x)=(x-1)^3+2~\text{.}\) Find formulas for:

  1. \(\displaystyle g(x)=\dfrac{1}{f(x)}\)
  2. \(\displaystyle h(x)=-f(x)\)
  3. \(\displaystyle j(x)=f^{-1}(x)\)
Answer
  1. \(\displaystyle g(x)=\dfrac{1}{(x-1)^3+2}\)
  2. \(\displaystyle h(x)=-(x-1)^3-2\)
  3. \(\displaystyle j(x)=1+\sqrt[3]{x-2}\)