Question:
Let $f$ and $g$ be defined as:
$$f(x) = 2x + 1, ~~~~x \in \mathbb{R}$$ $$g(x) = x^{2}, ~~~~~~~~~~~~x \in \mathbb{R}$$
Find
a) $~~f^{-1}(x)$
b) $~~f(g(x))$
c) $~~g(f(x))$
d) $~~f^{-1}(g(x))$
Attempted solutions:
The general idea I used was to set y = f(x) and then solve for x and that is my inverse function (if an inverse exists). For compound functions, I replaced x in the outer function with the inner function and then simplified. Since both f and g are defined on all of $\mathbb{R}$, I conclude that all combinations are also defined on all of $\mathbb{R}$ (especially since there are no fractions).
a)
$$y = 2x + 1 \Leftrightarrow x = \frac{y - 1}{2}$$
$$f^{-1}(x) = \frac{x-1}{2}, ~~~~~~x \in \mathbb{R}$$
b)
$$f(g(x)) = 2(x^2) + 1 = 2x^2 + 1, ~~~~~~x \in \mathbb{R}$$
c) $$g(f(x)) = (2x+1)^2 = 4x^2 + 4x +1, ~~~~~~x \in \mathbb{R}$$
d)
I attempted to get the inverse function for the answer in b):
$$y = 2x^2 + 1 \Leftrightarrow x = \sqrt{\frac{y - 1}{2}}$$
$$f^{-1}(g(x)) = \sqrt{\frac{x-1}{2}}, ~~~~~~x \in \mathbb{R}$$
However, the textbook answer is
$$f^{-1}(g(x)) = \frac{x^2-1}{2}, ~~~~~~x \in \mathbb{R}$$
Why is this approach wrong and what are some more productive approaches to solve this question?
$f(x)=2x+1$ so, as you wrote $f^{-1}(y)=\frac{y-1}{2}$. Being $g(x)=x^2$, you get $$ f^{-1}(g(x))=f^{-1}(x^2)=\frac{x^2-1}{2} $$ you have simply to substitute the expression for $g(x)$ inside the $f^{-1}(y)$.