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

Question

Quizplus · Accepted Answer

To find $f^{-1}(x)$, we solve for $x$ in terms of $y$: $y = e^{x-1} + 3$. Subtracting 3 and taking the natural logarithm of both sides gives $x = \ln(y-3) + 1$, so $f^{-1}(x) = \ln(x-3) + 1$. The domain of $f^{-1}(x)$ is $(3, \infty)$ because $x-3 > 0$ for the logarithm to be defined. The range of the original function $f(x)$ is $(3, \infty)$, which becomes the domain of $f^{-1}(x)$, and since $f(x)$ is continuous and increasing for all $x$, its inverse has a range of $(- \infty, \infty)$.

Given That f(x)=ex−1+3f ( x ) = e ^ { x - 1 } + 3f(x)=ex−1+3

Given That $f ( x ) = e ^ { x - 1 } + 3$