Question

use the given function f to answer parts (a) through (f) below. f(x)=ln(x + 3) b. there is no vertical asymptote. (d) find f⁻¹, the inverse of f. f⁻¹(x)=e^x - 3 (simplify your answer.) (e) find the domain and range of f⁻¹. the domain of f⁻¹ is (-∞,∞). (type your answer in interval - notation.) the range of f⁻¹ is (-3,∞). (type your answer in interval - notation.) (f) graph f⁻¹. choose the correct graph below.

Explanation:

Step1: Recall inverse - function domain and range relationship

The domain of the inverse function $f^{-1}(x)$ is the range of the original function $f(x)$, and the range of $f^{-1}(x)$ is the domain of $f(x)$.
For the function $f(x)=\ln(x + 3)$, the domain of $f(x)$ is $x+3>0$, i.e., $x>- 3$, so the domain of $f(x)$ is $(-3,\infty)$. The range of the natural - logarithm function $y = \ln(u)$ is $(-\infty,\infty)$.

Step2: Determine the domain and range of $f^{-1}(x)$

Since the domain of $f(x)$ is $(-3,\infty)$ and the range of $f(x)$ is $(-\infty,\infty)$, for $f^{-1}(x)=e^{x}-3$:
The domain of $f^{-1}(x)$ is the range of $f(x)$, so the domain of $f^{-1}(x)$ is $(-\infty,\infty)$.
The range of $f^{-1}(x)$ is the domain of $f(x)$, so the range of $f^{-1}(x)$ is $(-3,\infty)$.

Step3: Analyze the graph of $y = f^{-1}(x)=e^{x}-3$

The graph of $y = e^{x}$ is an exponential growth function that passes through the point $(0,1)$ with a $y$ - intercept of 1. The graph of $y=e^{x}-3$ is the graph of $y = e^{x}$ shifted down 3 units. It has a $y$ - intercept of $e^{0}-3=1 - 3=-2$ and a horizontal asymptote at $y=-3$.

Answer:

(e) Domain of $f^{-1}$: $(-\infty,\infty)$; Range of $f^{-1}$: $(-3,\infty)$
(f) The graph of $y = f^{-1}(x)$ is a curve that has a $y$ - intercept of - 2 and a horizontal asymptote at $y=-3$. Without seeing the actual graphs A, B, C, D in detail, but based on the characteristics, it should be a graph that starts just above $y = - 3$ (horizontal asymptote) and increases as $x$ increases, crossing the $y$ - axis at $y=-2$.