Finding the domain and range of a composite function f(g(x)) with fraction and trig I am trying to solve this problem. This is what I have done:
ATTEMPT 1: I have tried to think about the domain using composite functions but ended up no-where.
ATTEMPT 2: Restricting the domain for the f(x) and g(x)
ATTEMPT 3: Graphically; but this does not show any proof of finding the domaing and range
Question: Find the domain and range of f(g(x))
Given the following:
$$f(x)=\left(\frac{x^2+1}{2x^2-1}\right)$$
$$g(x)=2\pi-\arccos(x)$$
$$f(g(x))=2\pi-\arccos \left(\frac{x^2+1}{2x^2-1}\right)$$
$$f(x)=\frac{x^2+1}{2x^2-1}=\frac12+\frac3{2(2x^2-1)}.$$
As $2x^2-1\ge-1$, the range of $f(x)$ is $(\frac12,\infty)\cup(-\infty,-1]$.
So intersecting with the domain of the cosine, $g\left(\left(\dfrac12,1\right]\cup\{-1\}\right)=\left(\dfrac{5\pi}3,2\pi\right]\cup\{\pi\}$.
If $|2x^2 - 1| \leq 1$, then the fraction $\frac{x^2+1}{2x^2-1}$ has value outside $[-1,1]$ and hence the function is not defined. Thus the domain is $|x| \geq \sqrt{2}$.
When $|x| \geq \sqrt{2}$, the fraction takes all values in the range $\left(\frac{1}{2},1\right]$. Thus $arccos\left(\frac{x^2+1}{2x^2-1}\right)$ takes all values between $\frac{\pi}{3}$ and $0$. Thus the range is $\left(2\pi-\frac{\pi}{3}, 2\pi\right]$