What is a function, and what does $f(a)$ mean?

A function, written $f(x)$, is a rule that assigns exactly one output to every input. $f(a)$ means "substitute $a$ for every $x$" in that rule.

What are the domain and range of a function?

The domain is the set of inputs the function is allowed to take — you must exclude any value that would cause division by zero or the square root of a negative number. The range is the set of outputs the function can actually produce, found by considering how the function behaves across its whole domain.

For $g(x)=\dfrac{4}{x+2}$, the domain excludes $x=-2$ (written $x\neq-2$), since substituting it would divide by zero. For $h(x)=\sqrt{x-3}$, the domain requires $x-3\ge0$, so $x\ge3$, since a negative number has no real square root.

How do I find a composite function like $fg(x)$?

Work from the inside out: the function closest to $x$ is applied first, and its entire result becomes the input to the next function.

Given $f(x)=\dfrac{2x+3}{5}$ and $g(x)=x^2-1$, find $fg(2)$:

1. Apply $g$ first, since it's closest to $x$: $g(2)=2^2-1=3$.
2. Feed that result into $f$: $f(3) = \dfrac{2(3)+3}{5}=\dfrac{9}{5}$.

For $gf(x)$ with the same two functions, substitute the whole expression $f(x)$ into $g$:

How do I find the inverse of a function?

Write the function as $y=f(x)$, swap $x$ and $y$, then rearrange to make $y$ the new subject — this reverses the original mapping, so that $f^{-1}(f(x))=x$ for every valid $x$.

Swapping $x$ and $y$: $f^{-1}(x) = \dfrac{5x-3}{2}$.

How can a mapping diagram help me understand a function?

A mapping diagram draws each input on a left-hand list with an arrow pointing to its single output on a right-hand list. Because a function gives exactly one output per input, no input value ever has two arrows leaving it — if you see that on a diagram, the relationship shown is not a valid function.

Worked Example

Given $f(x)=3x-5$ and $g(x)=\dfrac{4}{x+2}$ for $x\neq-2$:

(a) Find $g^{-1}(x)$.

(b) Find $fg(3)$.

1. (a) Let $y=\dfrac{4}{x+2}$. Multiply both sides by $(x+2)$: $y(x+2)=4$.
2. Expand and rearrange to isolate $x$: $xy+2y=4 \implies xy=4-2y \implies x=\dfrac{4-2y}{y}$.
3. Swap $x$ and $y$: $g^{-1}(x) = \dfrac{4-2x}{x}$.
4. (b) Apply $g$ first: $g(3)=\dfrac{4}{3+2}=\dfrac{4}{5}$.
5. Feed into $f$: $f\left(\dfrac{4}{5}\right) = 3\left(\dfrac{4}{5}\right)-5 = \dfrac{12}{5}-5 = -\dfrac{13}{5}$.