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

Question

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

Quizplus · Accepted Answer

To find the inverse function of $f(x)$, we need to switch the places of $x$ and $y$ and solve for $y$: $$y=\frac{1}{x-3} \Rightarrow x=\frac{1}{y}-3 \Rightarrow y=\frac{1}{x+3}$$ Thus, $g(x)=\frac{1}{x+3}+3=3+\frac{1}{x}$.
The domain of $g$ is $(0,\infty)$ because $f(x)$ is defined only for $x>3$, and $g(x)$ is defined only for the values of $f(x)$ for which $x>3$. The range of $g$ is $(3,\infty)$ because $f(x)$ takes all values less than 0 as $x$ approaches 3 from the right, and $g(x)$ takes all values greater than 3 as $x$ approaches 0 from the right.
To find $g^t(a)$, we first need to find $g^{-1}(x)$: $$y=3+\frac{1}{x} \Rightarrow x=\frac{1}{y-3} \Rightarrow y=\frac{1}{x}+3=g^{-1}(x)$$
Then, we can use the property that $g^t(a)=(g^{-1})^t(a)$: $$g^t(a)=(g^{-1})^t(a)=\frac{1}{a}+3=\frac{1}{2}+3=\frac{7}{2}$$
Finally, we need to find the value of $-\frac{1}{g^t(a)}$: $$-\frac{1}{g^t(a)}=-\frac{1}{\frac{7}{2}}=-\frac{2}{7}$$
Therefore, the answer is A.

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

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