Quick questions on Composite and inverse functions: existence, formulas, domains and graphs

The composite $f \circ g$ is the function $(f \circ g)(x) = f(g(x))$: apply $g$ first, then $f$ to the result.

A function $f$ has an inverse (on its given domain) if and only if it is one-to-one: every value of $y$ in the range comes from exactly one $x$.

The inverse $f^{-1}$ "undoes" $f$: $f^{-1}(f(x)) = x$ for $x \in \text{dom}(f)$ and $f(f^{-1}(y)) = y$ for $y \in \text{range}(f)$.

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$. Point $(a, b)$ on $f$ corresponds to $(b, a)$ on $f^{-1}$. Horizontal asymptotes of $f$ become vertical asymptotes of $f^{-1}$ and vice versa.

For a non-one-to-one function $f$, choose a domain on which $f$ is one-to-one, then invert. Different restrictions give different inverses.

The inverse of a composition reverses the order:

Let $f(x) = \sqrt{x}$ (domain $x \ge 0$) and $g(x) = x - 4$. Find $f(g(x))$ and its domain.

$f(x) = 5 - 2 x$. Swap and solve: $x = 5 - 2 y$, so $y = \frac{5 - x}{2}$.

$f(x) = e^{x} + 1$ has domain $\mathbb{R}$ and range $(1, \infty)$.

$f(x) = (x - 1)^2$ on $\mathbb{R}$ is not one-to-one (the parabola has its vertex at $x = 1$ and is symmetric about it).

For $f(x) = 5 - 2 x$ and $f^{-1}(x) = \frac{5 - x}{2}$:

$f(g(x))$ applies $g$ first; $g(f(x))$ applies $f$ first. These give different functions in general.

$f^{-1}(x)$ means the inverse function, not $\frac{1}{f(x)}$. For example, if $f(x) = x + 2$, then $f^{-1}(x) = x - 2$, not $\frac{1}{x + 2}$.

Writing $y = \sqrt{x}$ as "the inverse of $y = x^2$" without specifying $x \ge 0$ misses half the picture: the original function must be made one-to-one first.

The domain of $f^{-1}$ equals the range of $f$. For $f(x) = e^x$ with range $(0, \infty)$, the inverse $\ln x$ has domain $(0, \infty)$.