# 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