Functions

A relation that assigns each input to exactly one output. Two parts: must act on every domain element, and must be well-defined (no input gets two outputs).

A graph represents a function iff every vertical line meets it in at most one point. Two intersections ⇒ one $x$ would map to two $y$s ⇒ not a function.

Set of allowed inputs. Natural domain = largest subset of $\mathbb{R}$ that gives real outputs.

Set of outputs actually achieved. Read it off the $y$-axis from the graph.

Denominator $\ne 0$ · radicand $\ge 0$ for even roots · argument $> 0$ for $\ln/\log$ · argument in $[-1,1]$ for $\arcsin/\arccos$. Intersect the allowed sets.

$\{x:x\ge 3\}$ ≡ $[3,\infty)$. $\{x:-2\le x<5\}$ ≡ $[-2,5)$. Inverted bracket means open endpoint.

(1) Write $y=f(x)$. (2) Swap $x$ and $y$. (3) Solve for $y$. That's $f^{-1}(x)$.

Graph of $f^{-1}$ = reflection of $f$ across the line $y=x$.

Iff $f$ is one-to-one. If $f$ fails the horizontal line test, restrict the domain to one side of the turning point first.

$D(f^{-1}) = R(f)$ and $R(f^{-1}) = D(f)$.

$x$ must be in $D(g)$ AND $g(x)$ must be in $D(f)$. Compose inner function first; then check the outer's domain restriction kicks in.

Test with $f(x)=2x$, $g(x)=x+1$: $(f\circ g)(1)=4$, $(g\circ f)(1)=3$.

$f^{-1} = f$. Examples: $y=x$, $y=-x$, $y=k/x$ (any $k$), $y=a-x$. Their graphs are symmetric about $y=x$ AND look the same swapped.

If no domain is stated, use the largest set giving real outputs. Always check denominators, radicals, logs.

$f^{-1}$ is the INVERSE function, NOT $1/f(x)$. They mean completely different things.

If you skip the one-to-one check, the 'inverse' you write down won't be a function. Always test (or restrict).