If $f(x)=e^x$, $g(x)=|x+2|$, and $h(x)=\frac{x-2}{x+1}$, what is the domain of $(f^{-1} \circ f^{-1})(x)$?
I believe the answer to be $(-\infty,\infty)$ but I am not exactly sure of how to prove such an answer. $(f^{-1} \circ f^{-1})(x)$ is equal to $\frac{1}{e^{1/e^x}}$. For any value of x, the composite function is never undefined and thus has a domain of $(-\infty,\infty)$. Please correct me if I am wrong.
The second question is: Evaluate $((fg)/h)(3)$. Show all work necessary to reach your conclusion.
Based on my calculations, the answer is $20e^3$. Please correct me if I am incorrect.
The last question is graph $(g \circ f)(x)$.
I am unsure of how to do so by hand without a graphing calculator accurately. If possible please provide an explanation of how to do so.
We know that $(g\circ f)(x) = |e^x + 2|$. You should use your understanding of domain and range to help you graph this.
First, the domain of this function is $(-\infty, \infty)$, since there are no restrictions on what values of $x$ we can put into $e^x$ or $|x|$. The range, however, is $[0, \infty)$ since the absolute value of anything has to be greater than or equal to 0.
Next, let's think about $g \circ f$ in terms of some elementary functions that you know. I assume you know what the function $e^x$ looks like. Then, $e^x + 2$ is a vertical shift of $e^x$ by 2 units. Finally, the absolute value will reflect any portion of the function that is negative about the $x$-axis.