Solve the problem.
-Given that $f ( x ) = e ^ { x - 1 } + 3$, find $f ^ { - 1 } ( x )$ and give the domain and range of $f ^ { - 1 } ( x )$
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \ln ( \mathrm { x } - 3 ) + 1$, domain $= ( 0 , \infty )$, range $= ( 0 , \infty )$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \ln ( x - 1 ) + 3$, domain $= ( - \infty , \infty )$, range $= ( - \infty , \infty )$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \ln ( \mathrm { x } - 1 ) + 3$, domain $= ( 3 , \infty )$, range $= ( - \infty , \infty )$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \ln ( x - 3 ) + 1$, domain $= ( 3 , \infty )$, range $= ( - \infty , \infty )$

Question

Quizplus · Accepted Answer

To find the inverse, solve $y = e^{x-1} + 3$ for $x$: $y - 3 = e^{x-1}$, then $x - 1 = \ln(y - 3)$, so $x = \ln(y - 3) + 1$. Thus, $f^{-1}(x) = \ln(x - 3) + 1$. The domain of $f^{-1}(x)$ is $x > 3$ (since $x - 3 > 0$), and the range is all real numbers.

Solve the Problem f(x)=ex−1+3f ( x ) = e ^ { x - 1 } + 3f(x)=ex−1+3

Solve the Problem $f ( x ) = e ^ { x - 1 } + 3$