Hey!
I've encountered many inverse trig function questions where they ask me to find the domain and range of a function within the function. I'm a little bit confused as to how to work these questions out. For example, this question:
If anyone could lend me a hand, I'd really appreciate it!!
\[ \text{Since }h(x) = \tan^{-1}x\text{ has natural domain all real }x,\\ \text{we only need to consider any domain restrictions on }g(x)=\sqrt{1-x^2}. \]
Which is of course, just a semi-circle with radius 1, so we know its domain is \( -1\leq x \leq 1 \). This will then be the domain of \( h(g(x)) = \tan^{-1} \sqrt{1-x^2} \).
\[ \text{Now, the }\textbf{range}\text{ of the inner function }g(x)\\ \text{will be treated as a }\textbf{domain}\text{ restriction}\\ \textbf{only}\text{ for the outer function }h(x). \]
\[\text{Here, }\sqrt{1-x^2}\text{ has range }0\leq y \leq 1.\\ \text{So we need to consider the range of }\tan^{-1}x\\ \textbf{when}\text{ we restrict the range of }\tan^{-1}x\text{ to }0\leq x \leq 1. \]
\[ \text{Since }h(x) = \tan^{-1}x\text{ is monotonic increasing (just look at its graph)}\\ \text{we can basically infer it from the graph by focusing on its endpoints.}\\ \text{We know that }\tan^{-1}0 = 0\text{ and }\tan^{-1}1 = \frac\pi4\\ \text{so therefore the range will be }\boxed{0\leq y \leq \frac\pi4}. \]