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Assume that $f$ is one $-$ to $-$ one function
(a) If $f(6)=17$ what is $f^{-1}(17)$
(b) If $f^{-1}(3)=2$ what is $f(2)$

(a) $f(6)=17 \rightarrow f^{-1}(17)=? ?$

$f^{-1}(17)=6$

(b) $f^{-1}(3)=2 \rightarrow f(2)=? ?$

$f(2)=3$

Find a formula for the inverse of the function

$f(x)=\frac{4 x-1}{2 x+3}$

(1) $y=\frac{4 x-1}{2 x+3}$

(2) $2 y x+3 y=4 x-1$

$2 y x-4 x=-3 y-1$

$x(2 y-4)=-(3 y+1) \quad \div 2 y-4$

$x=\frac{-(3 y+1)}{2 y-4}$

(3) $y=\frac{-(3 x+1)}{2 x-4}$

(4) $f^{-1}(x)=\frac{-(3 x+1)}{2 x-4}$

Let $f(x)=\ln \left(e^{x}-1\right)$
(a) Find the domain of $f$
(b) Find $F^{-1}$ and its domain

(a) $f(x)=\ln \left(e^{x}-1\right)$

$e^{x}-1>0$

$e^{x}>1$

$\ln e^{x}>\ln (1)$

$x>0$

Domain $(0,+\infty)$

(b) $F^{-1}=? ?$

$f(x)=\ln \left(e^{x}-1\right)$

$y=\ln \left(e^{x}-1\right)$

$e^{y}=e^{\ln \left(e^{x}-1\right)}$

$e^{y}=e^{x}-1$

$e^{x}=e^{y}+1$

$\ln e^{x}=\ln \left(e^{y}+1\right)$

$x=\ln \left(e^{y}+1\right)$

$y=\ln \left(e^{x}+1\right)$

$F^{-1}(x)=\ln \left(e^{x}+1\right)$

$e^{x}+1>0$

$e^x>-1$

$\ln e^{x} > \ln-1$

$x>\ln (-1)$

Drmain $(-\infty,+\infty)$

Let $f(x)=\frac{x}{1+2 x}$
(a) Find the domain of $f$
(b) Find $F^{-1}(x)$
(c) Find the range of $F$

(a) $f(x)=\frac{x}{1+2 x}$

$1+2 x=0 \quad 2 x=-1 \quad x=-\frac{1}{2}$

Domain $=R /\left\{-\frac{1}{2}\right\}$

(b) $F^{-1}(x) ?$

$y=\frac{x}{1+2 x}$

$y+2 y x=x$

$2 y x-x=-y$

$x(2 y-1)=-y$

$x=\frac{-y}{2 y-1}$

$y=\frac{-x}{2 x-1}$

$F^{-1}(x)=\frac{-x}{2 x-1}$

$D F^{-1}=R /\left\{\frac{1}{2}\right\}$