Department of Mathematics



Functions and Their Inverses



Self-Test:


1)       If the point $(2,-3)$ is on the graph of $y=f(x),$ then the graph of $y=f^{-1}(x)$ must contain the point:

      $(3,-2)$
      $\displaystyle{ \left(\frac{-1}{2}, \frac{1}{3} \right) }$
      $\displaystyle{ \left(\frac{-3}{2}, \frac{2}{3} \right) }$
      $(-3,2)$
      $(-2,3)$

Hint If the functions $f(x)$ and $f^{-1}(x)$ are inverses, and the point $(c,d)$ is on the graph of $f(x),$ the the point $(d,c)$ is on the graph of $f^{-1}(x).$

2)       If $f(2)=5, \; f(5)=-1, \; g(5) = 4, \; g(-1) = 5,\;$ then $g^{-1}\left( f \left( f^{-1}(f(2)) \right) \right) = $

      $-1$
      $4$
      $5$
      $2$
      Not Defined

Hint If the functions $f(x)$ and $f^{-1}(x)$ are inverses, then $f(f^{-1}(x)) = f^{-1}(f(x)) = x.$ This is known as the Cancellation Property of Inverses

3)       If $\displaystyle{ f(x) = \frac{x+2}{3x-1} }$ then $f^{-1}(x) = $

      $\displaystyle{ \frac{x+2}{3x-1} }$
      $\displaystyle{ \frac{3x-1}{x+2} }$
      $\displaystyle{ \frac{3-x}{2x} }$
      $\displaystyle{ \frac{x-1}{3x+2} }$
      $\displaystyle{ \frac{3x+2}{x-1} }$

Hint If the functions $f(x)$ and $f^{-1}(x)$ are inverses, then their domain (input: valid $x$-values) and their range (output: $y$-values) switch! So, simply switch the roles of $x$ and $y$ and solve for the "new $y.$"

4)       Let $\displaystyle{f(x) = \frac{1}{x-3} }$ and $g(x)=\sqrt{x}.$ Then for $(f \circ g)(x),$

      Domain $= \{ x \in \mathbb{R} \; \big| \; x>0 \text{ and } x \neq 3 \}$     Range = $ \{ y \in \mathbb{R} \; \big| \; y > 0$ or $y < -\frac13 \}$
      Domain $= \{ x \in \mathbb{R} \; \big| \; x>0 \text{ and } x \neq 3 \}$     Range = $ \{ y \in \mathbb{R} \; \big| \; y \neq 0 \}$
      Domain $= \{ x \in \mathbb{R} \; \big| \; x>3 \}$     Range = $ \{ y \in \mathbb{R} \; \big| \; y \neq 0 \}$
      Domain $= \{ x \in \mathbb{R} \; \big| \; x \geq 0 \text{ and } x \neq 9 \}$     Range = $ \{ y \in \mathbb{R} \; \big| \; y > 0$ or $ y<-\frac13 \}$
      Domain $= \{ x \in \mathbb{R} \; \big| \; x \geq 0 \text{ and } x \neq 9 \}$     Range = $ \{ y \in \mathbb{R} \; \big| \; y\neq> 0 \}$

Hint 1 $(f \circ g)(x) = f(g(x))$ by definition
Hint 2 The domain of $f \circ g$ must satisfy both the domains of $g$ and $f \circ g$

5)       If $(a,b)$ is on the graph of both $f(x)$ and its inverse, $f^{-1}(x),$ then:

      either $a=0$ or $b=0$
      $a=b$
      $f(x)=f^{-1}(x)$ for all values of $x$
      the graph of $y=f(x)$ is symmetric about a line
      $a=f(f(a))$

Hint See the hint from #1

6)       If $\log_5 125 = 3$ means that $5^3 = 125,$ then $\log_5 \frac{1}{125} = $

      $\frac{1}{3}$
      $-\frac{1}{3}$
      $-3$
      $3$
      undefined


7)       Which of the following statements is true?
      The graph of $y=e^x$ crosses the $x$-axis exactly once
      The graph of $y=\ln(x)$ gets infinitely close to the graph $y=0$ as $x$ goes to positive infinity
      If $f(x)=e^x$ and $g(x)= \ln(x),$ then $f(g(x)) = g(f(x)) = 1$
      The graphs of $y=e^x$ and $y=\ln(x)$ intersect exactly once
      The range of the function $f(x)=e^x$ is equal to the domain of $g(x) = \ln(x)$ which is equal to the interval $(0, +\infty)$

Hint Try creating a rough sketch of the graphs of $y=e^x$ and $y=\ln(x) = \log_e x,$ in order to observe the properties more easily

8)       The solution to $2\ln(x)+\ln(x+2) - \ln(x^2+2x) = -\ln(2)$ is

      $2$
      $-2$
      $-\ln(2)$
      $-\frac{1}{2}$
      $\frac{1}{2}$

Hint 1 Use the following properties of logarithms:

1. $\log_b(r \cdot s) = \log_b r + \log_b s $
2. $\log_b( \frac{r}{s}) = \log_b r - \log_b s $
3. $\log_b( r^{s}) = s \cdot \log_b r $
Hint 2 If $\log_b (c) = s \cdot \log_b (d),$ then we can conclude that $c=d$ since logarithmic functions are one-to-one

9)       $\frac{1}{2}$ is NOT equal to

      $e^{\ln 0.5}$
      $\ln 1 - \ln 2$
      $e^{-\ln 2}$
      $\frac{3 e^0}{6}$
      $\sqrt{e^{-\ln 4}}$

Hint Use the fact that $y=e^x$ and $y=\ln x = \log_e x$ are inverses. Then apply hints from #1 and #8.

10)       $16^{x+3} = \frac{1}{8^{x+1}}$ implies that $x=$

      $0$
      $-6$
      $-\frac{4}{7}$
      $-\frac{15}{7}$
      undefined

Hint 1 Try expressing each side of the equal sign with the same base
(using properties of exponents when needed)
Hint 2 If $b^c = b^d,$ then we can conclude that $c=d$ since exponential functions are one-to-one





Worked Examples and Practice Problems