Inverse Functions

1. For the following functions, find $f^{-1}(x)$ if it exists. If not, explain why.

1. $\displaystyle{f(x) = \sqrt[3]{\frac{2x+1}{3-x}}}$   $\ans{ \begin{array}{l} \textrm{For any point } (x,y) \textrm{ on the graph of } y=f(x)\textrm{, we know}\\ \textrm{that }y=\sqrt[3]{(2x+1)/(3-x)}\\\\ \textrm{Consequently, for any point } (x,y) \textrm{ on}\\ \textrm{the graph of } y=f^{-1}(x), \textrm{ we know that } x=\sqrt[3]{(2y+1)/(3-y)}.\\\\ \textrm{Thus,}\\\\ \begin{array}{rcl} x &=& \displaystyle{\sqrt[3]{\frac{2y+1}{3-y}}}\\ x^3 &=& \displaystyle{\frac{2y+1}{3-y}}\\ 3x^3 - x^3 y &=& 2y + 1\\ 3x^3 - 1 &=& 3y + x^3 y\\ 3x^3 - 1 &=& y(2+x^3)\\ y &=& \displaystyle{\frac{3x^3 - 1}{2+x^3}} \end{array}\\\\ \textrm{Hence, the inverse function exists and } f^{-1} = \displaystyle{f^{-1}(x) = \frac{3x^3 - 1}{2+x^3}}\\\\ \end{array}}$

2. $\displaystyle{f(x) = \sqrt[5]{(2x-3)^{3/7} + 4}}$   $\ans{\displaystyle{f^{-1}(x) = \frac{(x^5-4)^{7/3} + 3}{2}}\\\\}$

3. $\displaystyle{f(x) = -3 \left| -2x+1 \right| + 7}$   $\ans{\textrm{no inverse as$x=0$and$x=1$produce the same$y$-value}}$

4. $\displaystyle{f(x) = \frac{x+1}{x-2}}$   $\ans{\displaystyle{f^{-1}(x) = \frac{2x+1}{x-2}}}$

5. $\displaystyle{f(x) = \frac{-2x-5}{x+2}}$   $\ans{\displaystyle{f^{-1}(x) = \frac{-2x-5}{x+2}}}$

6. $\displaystyle{f(x) = \sqrt[3]{\frac{x-2}{5}}}$   $\ans{\displaystyle{f^{-1}(x) = 5x^3 +2}}$

2. For each of the following, draw the graph of the function and then determine, based on the graph, whether the function has an inverse or not.

1. $\displaystyle{f(x) = x^2-3x}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{note, two$x$-intercepts at$(0,0)$and$(3,0)$so it}\\ \textrm{fails the horizontal line test - thus, there is no inverse function} \end{array}}}$

2. $\displaystyle{f(x) = |x-2| + 1}$   $\ans{\displaystyle{ \begin{array}{l} \textrm{note,$f(0)=2$and$f(4)=2$, so it fails the horizontal}\\ \textrm{line test - thus, there is no inverse function} \end{array}}}$

3. $\displaystyle{f(x) = \sqrt[3]{x} + 2}$   $\ans{\displaystyle{f^{-1}(x) = (x-2)^3}}$

4. $\displaystyle{f(x) = \frac{1}{x}}$   $\ans{\displaystyle{f^{-1}(x) = \frac{1}{x}}}$