Functions and Graphs

Functions and graphs form one of the most heavily tested topics in H2 Mathematics (9758), underpinning everything from differentiation to integration and complex numbers. A firm grasp of domain and range, the conditions for inverses and composites to exist, and the rules for graph transformations will save marks across the entire paper.

What you need to know

• Domain and range: The domain is the set of input values; the range is the set of output values. Always state both when defining a function, e.g. $f(x) = \sqrt{x-1}$, $x \geq 1$.
• One-one functions and the horizontal-line test: A function $f$ has an inverse $f^{-1}$ if and only if $f$ is one-one (injective) — every horizontal line meets the graph at most once. If $f$ is not one-one, restrict the domain until it is.
• Finding $f^{-1}$: Write $y = f(x)$, rearrange for $x$ in terms of $y$, then swap notation. The domain of $f^{-1}$ equals the range of $f$, and vice versa. The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$.
• Composite functions $fg$: The composite $fg$ exists if and only if the range of $g$ is a subset of the domain of $f$, i.e. $R_g \subseteq D_f$. Evaluate right-to-left: $fg(x) = f(g(x))$.
• Graph transformations: For a graph $y = f(x)$: replacing $x$ with $x - a$ translates by $a$ units in the positive $x$-direction; replacing $y$ with $y - b$ (i.e. adding $b$) translates by $b$ units upward; replacing $x$ with $\frac{x}{a}$ stretches horizontally by factor $a$; multiplying by $a$ stretches vertically by factor $a$; replacing $x$ with $-x$ reflects in the $y$-axis.
• Modulus graphs: $y = |f(x)|$ reflects any part of $y = f(x)$ that lies below the $x$-axis upward. $y = f(|x|)$ keeps the right half of $y = f(x)$ and reflects it in the $y$-axis (the left half is discarded).

Worked example 1 — Finding an inverse function and its domain

Problem. The function $f$ is defined by $f(x) = 2x + 3$, $x \in \mathbb{R}$, $x \geq 0$. Find $f^{-1}(x)$ and state its domain.

Solution.

$f$ is one-one on $x \geq 0$ (it is strictly increasing), so $f^{-1}$ exists.

Let $y = 2x + 3$ and solve for $x$:

Therefore

The domain of $f^{-1}$ equals the range of $f$. Since $x \geq 0$, we have $f(x) = 2x + 3 \geq 3$, so the range of $f$ is $[3, \infty)$.

Domain of $f^{-1}$: $x \geq 3$.

Worked example 2 — Composite functions

Problem. Functions $g$ and $h$ are defined by $g(x) = x^2 + 1$, $x \in \mathbb{R}$, and $h(x) = \ln x$, $x > 0$. Determine whether $hg$ exists, and if so find $hg(x)$ and state its range.

Solution.

For $hg$ to exist we need $R_g \subseteq D_h$.

• Range of $g$: since $x^2 \geq 0$, we have $g(x) = x^2 + 1 \geq 1$, so $R_g = [1, \infty)$.
• Domain of $h$: $D_h = (0, \infty)$.

Since $[1, \infty) \subseteq (0, \infty)$, the composite $hg$ exists.

For the range: $x^2 + 1 \geq 1$, so $\ln(x^2 + 1) \geq \ln 1 = 0$. As $|x| \to \infty$, $\ln(x^2+1) \to \infty$. The value $\ln 1 = 0$ is attained at $x = 0$.

Range of $hg$: $[0, \infty)$.

Worked example 3 — Modulus graph transformation

Problem. The diagram below is not required; describe how to sketch $y = |f(x)|$ and $y = f(|x|)$ given $f(x) = x - 2$.

Solution.

First sketch $y = x - 2$, a straight line with gradient 1 crossing the $x$-axis at $x = 2$.

$y = |x - 2|$: The portion of the line where $y < 0$ (i.e. $x < 2$) is reflected upward in the $x$-axis. The result is a V-shape with vertex at $(2, 0)$, with gradient $-1$ for $x < 2$ and gradient $1$ for $x > 2$.

$y = f(|x|) = |x| - 2$: Replace $x$ with $|x|$. Keep the graph for $x \geq 0$ unchanged (a line with gradient 1 from $(0,-2)$ upward), then reflect it in the $y$-axis for $x < 0$. The result is a V-shape with vertex at $(0, -2)$, crossing the $x$-axis at $x = \pm 2$.

Common mistakes

• Forgetting to state the domain of $f^{-1}$: writing only the rule for $f^{-1}$ without noting that its domain equals the range of $f$ loses a mark in almost every inverse question.
• Getting the condition for $fg$ backwards: the condition is $R_g \subseteq D_f$ (range of the inner function inside domain of the outer), not $R_f \subseteq D_g$.
• Applying $y = f(|x|)$ incorrectly: students often reflect the whole graph instead of keeping the right half and reflecting it — remember the left half of the original is discarded, not retained.
• Horizontal vs vertical stretch confusion: a factor of $a$ inside $f$, e.g. $f(ax)$, compresses the graph horizontally by factor $a$ (not stretches); only $f\!\left(\frac{x}{a}\right)$ stretches it by factor $a$.