Unit 2 · Functions and Their Properties　函数及其性质

By the end of this chapter you can:

1. State the definition of a function, find its domain and range, and represent it as a mapping
2. Test whether a function is even, odd, or neither and describe the symmetry consequence
3. Define monotonicity formally and use it to compare function values without a calculator
4. Find inverse functions, compose two functions, and determine whether two differently-written functions are identical

Exam weight on past CSCA papers: ~12% (5–6 of 48 MCQs).

2.1　Function Definition and Domain　函数定义与定义域

A function $f$ from set $A$ to set $B$ is a rule that assigns to every element $x \in A$ exactly one element $y \in B$. We write $y = f(x)$.

• Domain (定义域): the set $A$ — all permissible inputs.
• Codomain (上域): the set $B$ — declared target.
• Range (值域): the subset of $B$ actually produced, $\{f(x) : x \in A\} \subseteq B$.

The domain is determined by natural restrictions on the formula:

When multiple restrictions apply, the domain is their intersection.

🔑 Domain strategy: List every restriction separately, then intersect. For $f(x) = \dfrac{\sqrt{x-1}}{x-3}$ you get $x - 1 \ge 0$ AND $x - 3 \ne 0$, giving domain $[1, 3) \cup (3, +\infty)$.

Worked Example 2.1.A

Find the domain and range of $f(x) = \sqrt{4 - x^{2}}$.

Solution.

Domain: The radicand must be non-negative.
$$4 - x^{2} \ge 0 \implies x^{2} \le 4 \implies -2 \le x \le 2.$$
Domain: $[-2, 2]$.

Range: For $x \in [-2, 2]$, the radicand $4 - x^{2}$ ranges from $0$ (at $x = \pm 2$) to $4$ (at $x = 0$). Taking the square root:
$$f(x) = \sqrt{4 - x^{2}} \in [0, 2].$$

$$\boxed{\ \text{Domain} = [-2,\ 2],\quad \text{Range} = [0,\ 2]\ }$$

⚠️ Range ≠ codomain. The codomain of $f: \mathbb{R} \to \mathbb{R}$ is all of $\mathbb{R}$, but the range of $f(x) = x^{2}$ is only $[0, +\infty)$. CSCA questions often test this distinction when asking "what is the range of …?"

2.2　Even and Odd Functions　函数的奇偶性

Before testing parity, confirm the domain is symmetric about 0 (i.e. $x \in$ domain $\Rightarrow -x \in$ domain). If the domain is not symmetric, the function is neither even nor odd by definition.

Immediate consequences:

• If $f$ is odd and $0$ is in the domain, then $f(0) = 0$. (Plugging $x = 0$: $f(0) = -f(0)$, so $f(0) = 0$.)
• The sum of two even functions is even; the sum of two odd functions is odd.
• The product of two odd functions is even.

Worked Example 2.2.A

Determine whether $f(x) = x^{3} - x$ is even, odd, or neither.

Solution.

Step 1 — Domain check. $f$ is a polynomial, so domain $= \mathbb{R}$. Since $\mathbb{R}$ is symmetric about $0$, parity testing applies.

Step 2 — Compute $f(-x)$.
$$f(-x) = (-x)^{3} - (-x) = -x^{3} + x = -(x^{3} - x) = -f(x).$$

Since $f(-x) = -f(x)$ for all $x \in \mathbb{R}$, the function is odd.

$$\boxed{\ f(x) = x^{3} - x \text{ is an odd function.}\ }$$

⚠️ Common trap: Students test parity on $f(x) = \dfrac{1}{x} + x$ and get odd, then plug in a non-symmetric domain example like $f$ on $[1, 3]$ and wonder why. Always state the domain first — on $[1, 3]$ the function has no parity at all.

🔑 Quick shortcut: A function built entirely from odd powers of $x$ (and no constant term) is odd. Built entirely from even powers (and possibly a constant) is even. Mixed powers → neither. E.g. $x^{4} + x^{2} - 3$ is even; $x^{5} + x$ is odd; $x^{2} + x$ is neither.

2.3　Monotonicity　函数的单调性

Formal definition: $f$ is increasing on interval $I$ if for all $x_{1}, x_{2} \in I$ with $x_{1} < x_{2}$, we have $f(x_{1}) < f(x_{2})$. It is decreasing on $I$ if $x_{1} < x_{2} \Rightarrow f(x_{1}) > f(x_{2})$.

A function can be increasing on one interval and decreasing on another. We always name the interval when stating monotonicity.

Key properties of monotonic functions on the same interval $I$:

🔑 Monotonicity + inequality shortcut: If $f$ is increasing and $f(a) < f(b)$, then $a < b$ — you can "pull back" the inequality through the function. This is how CSCA questions like "solve $f(x) > f(3)$" are answered without finding $f^{-1}$.

Worked Example 2.3.A

The function $f(x)$ is defined on $\mathbb{R}$ and is monotonically increasing. Given that $f(2m - 1) < f(3 - m)$, find the range of $m$.

Solution.

Because $f$ is increasing, $f(u) < f(v) \Leftrightarrow u < v$. Therefore:
$$2m - 1 < 3 - m.$$
$$3m < 4 \implies m < \dfrac{4}{3}.$$

$$\boxed{\ m < \dfrac{4}{3}\ }$$

⚠️ Trap: If the problem said "monotonically decreasing," the inequality flips: $f(u) < f(v) \Leftrightarrow u > v$. Check the direction of monotonicity before pulling back.

2.4　Inverse Functions　反函数

A function $f: A \to B$ has an inverse $f^{-1}: B \to A$ if and only if $f$ is one-to-one (injective) — no two different inputs give the same output.

How to find $f^{-1}$:

1. Write $y = f(x)$.
2. Solve for $x$ in terms of $y$.
3. Swap labels: replace $x$ with $y$ and $y$ with $x$ (so the inverse is written with the standard variable $x$).
4. State the domain of $f^{-1}$ = range of $f$.

Key fact: The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$.

Composition identity: $f^{-1}(f(x)) = x$ for $x \in$ domain of $f$, and $f(f^{-1}(x)) = x$ for $x \in$ domain of $f^{-1}$.

Monotonicity is preserved: If $f$ is increasing (decreasing) on its domain, then $f^{-1}$ is also increasing (decreasing) on its domain.

Worked Example 2.4.A

Find the inverse of $f(x) = 2x - 3$ and verify by composition.

Solution.

Step 1: Write $y = 2x - 3$.

Step 2: Solve for $x$:
$$y + 3 = 2x \implies x = \dfrac{y + 3}{2}.$$

Step 3: Swap variables:
$$f^{-1}(x) = \dfrac{x + 3}{2}.$$

Step 4: Domain of $f^{-1}$ = range of $f$ = $\mathbb{R}$.

Verification:
$$f(f^{-1}(x)) = 2 \cdot \dfrac{x+3}{2} - 3 = (x + 3) - 3 = x. \checkmark$$

$$\boxed{\ f^{-1}(x) = \dfrac{x+3}{2},\quad \text{domain } \mathbb{R}\ }$$

⚠️ The graph reflection trap: A question may give you the graph of $f$ and ask you to sketch $f^{-1}$. The reflection is across $y = x$, not across the $x$-axis or $y$-axis. Reflecting across the $x$-axis gives $-f(x)$; across the $y$-axis gives $f(-x)$.

🔑 Quick domain check: Always double-check that the domain of $f^{-1}$ equals the range of $f$ (not the domain of $f$). Students often copy the wrong interval.

2.5　Same Function Determination　判断同一函数

Two functions $f$ and $g$ are identical (the same function) if and only if both conditions hold:

1. Same domain: $D(f) = D(g)$.
2. Same rule: $f(x) = g(x)$ for all $x$ in that shared domain.

Even if the formulas simplify to the same expression, the functions differ if their domains differ.

⚠️ The cancellation trap: Writing $\dfrac{x^2 - 1}{x - 1} = x + 1$ is a valid algebraic step, but the resulting function $x + 1$ is not the same function as the original — you created a new function by removing a hole. CSCA examiners exploit this distinction every year.

Worked Example 2.5.A

Are $f(x) = \ln(x^{2})$ and $g(x) = 2\ln x$ the same function?

Solution.

Domain of $f$: $x^{2} > 0$, i.e. $x \ne 0$. So $D(f) = \mathbb{R} \setminus \{0\} = (-\infty, 0) \cup (0, +\infty)$.

Domain of $g$: $x > 0$. So $D(g) = (0, +\infty)$.

Since $D(f) \ne D(g)$ ($f$ is defined for negative $x$ but $g$ is not), the two functions are not identical.

(Note: on the common domain $(0, +\infty)$, both give $2 \ln x$, but the differing domains disqualify identity.)

$$\boxed{\ f \text{ and } g \text{ are NOT the same function.}\ }$$

2.6　Composite Functions　复合函数

Given two functions $f$ and $g$, the composite function $(f \circ g)(x) = f(g(x))$ applies $g$ first, then $f$ to the result.

Domain of $f \circ g$:
$$D(f \circ g) = \{x \in D(g) : g(x) \in D(f)\}.$$
In words: $x$ must be in the domain of $g$, AND the output $g(x)$ must land in the domain of $f$.

⚠️ Order matters: $f \circ g \ne g \circ f$ in general. The notation $f(g(x))$ means "$g$ acts first." Many students reverse the order — always read right to left: the function closest to $x$ acts first.

Monotonicity of compositions (same-sign rule):

🔑 Mnemonic — "same signs give increasing": If $f$ and $g$ have the same monotone direction (both ↑ or both ↓), then $f \circ g$ is increasing. If they differ (one ↑, one ↓), $f \circ g$ is decreasing. Like multiplying signs: $+\times+ = +$, $-\times- = +$, etc.

Worked Example 2.6.A

Let $f(x) = \sqrt{x - 1}$ and $g(x) = x^{2} - 2$.

(a) Find $(f \circ g)(x)$ and state its domain.
(b) Find $(g \circ f)(x)$ and state its domain.

Solution.

(a) $f(g(x)) = f(x^{2} - 2) = \sqrt{(x^{2} - 2) - 1} = \sqrt{x^{2} - 3}$.

Domain: Need $x^{2} - 3 \ge 0$, i.e. $x^{2} \ge 3$, so $x \le -\sqrt{3}$ or $x \ge \sqrt{3}$.
$$D(f \circ g) = (-\infty,\ -\sqrt{3}\,] \cup [\sqrt{3},\ +\infty).$$

(b) $g(f(x)) = g(\sqrt{x-1}) = (\sqrt{x-1})^{2} - 2 = (x - 1) - 2 = x - 3$.

Domain: First, $f$ requires $x - 1 \ge 0$, i.e. $x \ge 1$. The output $\sqrt{x-1} \ge 0$, which is always in the domain of $g$. So:
$$D(g \circ f) = [1,\ +\infty).$$

$$\boxed{\ f(g(x)) = \sqrt{x^{2}-3},\ D = (-\infty,-\sqrt{3}\,]\cup[\sqrt{3},+\infty);\ \quad g(f(x)) = x - 3,\ D = [1,+\infty)\ }$$

Try it!　自测练习

Q1. Find the domain of $f(x) = \dfrac{\sqrt{x + 2}}{x - 1}$.

Q2. Determine whether $h(x) = x \cdot \sin x$ is even, odd, or neither. (You may use the fact that $\sin(-x) = -\sin x$.)

Q3. The function $f$ is decreasing on $\mathbb{R}$. Given $f(3 - 2m) > f(m + 1)$, find the range of $m$.

Q4. Find the inverse of $g(x) = \dfrac{x}{x - 2}$ (for $x \ne 2$), and state its domain.

Q5. Let $f(x) = x + 1$ and $g(x) = x^{2}$. Find $(f \circ g)(3)$ and $(g \circ f)(3)$.

📌 Chapter summary

Topic	Key skill
2.1 Domain & Range	List each restriction, intersect; range requires tracking outputs
2.2 Even & Odd	Check domain symmetry first; test $f(-x)$ vs $\pm f(x)$; odd $\Rightarrow f(0) = 0$
2.3 Monotonicity	Formal definition with $x_1 < x_2$; use monotonicity to "pull back" inequalities
2.4 Inverse Functions	Swap $x \leftrightarrow y$; domain of $f^{-1}$ = range of $f$; graph reflects over $y = x$
2.5 Same Function	Identical iff same domain AND same rule; cancelling holes creates a new function
2.6 Composite Functions	$f(g(x))$: $g$ acts first; domain = inputs where $g(x) \in D(f)$; monotonicity: same-sign $\Rightarrow$ increasing

What's next → Unit 3 applies everything here to specific elementary functions: power, exponential, logarithmic, and quadratic — you'll use domain rules, parity, and monotonicity on each one.