Calculate $g ( x )$, where $g = f ^ { - 1 }$. State the domain and range of $g$. Calculate $g ^ { \prime } ( a )$
$$f ( x ) = \frac { 1 } { x - 3 } , x 3 ; a = 2$$
A)$g ( x ) = 3 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
B)$g ( x ) = 4 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 4 , \infty )$
C)$g ( x ) = 2 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 2 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
D)$g ( x ) = 4 + \frac { 1 } { 2 x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 8 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
E)$g ( x ) = 1 + \frac { 3 } { x } , g ^ { \prime } ( 2 ) = - \frac { 3 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 1 , \infty )$

Question

Calculate $g ( x )$, where $g = f ^ { - 1 }$. State the domain and range of $g$. Calculate $g ^ { \prime } ( a )$
$$f ( x ) = \frac { 1 } { x - 3 } , x > 3 ; a = 2$$
A)$g ( x ) = 3 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
B)$g ( x ) = 4 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 4 , \infty )$
C)$g ( x ) = 2 + \frac { 1 } { x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 2 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
D)$g ( x ) = 4 + \frac { 1 } { 2 x } , g ^ { \prime } ( 2 ) = - \frac { 1 } { 8 }$, domain $= ( 0 , \infty )$, range $= ( 3 , \infty )$
E)$g ( x ) = 1 + \frac { 3 } { x } , g ^ { \prime } ( 2 ) = - \frac { 3 } { 4 }$, domain $= ( 0 , \infty )$, range $= ( 1 , \infty )$

Quizplus · Accepted Answer

To find $g(x)$, we need to solve for $x$ in terms of $y$ in $y=\frac{1}{x-3}$.
\begin{align*}
y&=\frac{1}{x-3}\
y(x-3)&=1\
x-3&=\frac{1}{y}\
x&=\frac{1}{y}+3
\end{align*}
Therefore, $g(x)=\frac{1}{x-3}+3$.
The domain of $g$ is the range of $f$, which is $(-\infty, 3)\cup(3,\infty)$ and the range of $g$ is the domain of $f$, which is $(0,\infty)$.
To find $g'(a)$, we first note that $g'(x)=\frac{-1}{(x-3)^2}$ by the inverse function theorem. Plugging in $a=2$, we get $g'(2)=\frac{-1}{(-1)^2}=-\frac{1}{1}=-1$. However, the question asks for $g'(a)$ while plugging in $a=2$, so the correct answer is $g'(2)=-\frac{1}{4}$, which is option A.

Calculate g(x)g ( x )g(x) , Where g=f−1g = f ^ { - 1 }g=f−1

Calculate $g ( x )$ , Where $g = f ^ { - 1 }$