Grade 12 Advanced Functions (MHF4U)
MHF4U is the gateway to calculus. It demands real fluency with functions — not just procedural skill, but the ability to read a function's structure and predict its behaviour. This guide covers all four strands in the Ontario curriculum: polynomial and rational functions, exponential and logarithmic functions, advanced trigonometry, and the structural properties of functions (composition and inverses). For each topic, worked examples build from simple entry points to exam-level complexity before the practice set tests your mastery.
1. Polynomial Functions
A polynomial function of degree $n$ has the form $f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ where $a_n \neq 0$. The leading term $a_n x^n$ controls the big-picture behaviour; the roots and their multiplicities control the local behaviour near each zero.
End Behaviour Summary
- $a_n > 0$, $n$ even: both ends up ($\nearrow \cdots \nearrow$)
- $a_n < 0$, $n$ even: both ends down ($\searrow \cdots \searrow$)
- $a_n > 0$, $n$ odd: left end down, right end up ($\searrow \cdots \nearrow$)
- $a_n < 0$, $n$ odd: left end up, right end down ($\nearrow \cdots \searrow$)
Worked Example 1.1 — Reading End Behaviour
Without graphing, describe the end behaviour of each function.
(a) $f(x) = -2x^5 + 3x^3 - x$
Leading term: $-2x^5$. Degree 5 (odd), leading coefficient $-2$ (negative). As $x \to +\infty$, $f(x) \to -\infty$; as $x \to -\infty$, $f(x) \to +\infty$. Left end up, right end down.
(b) $g(x) = 4x^4 - 7x^2 + 1$
Leading term: $4x^4$. Degree 4 (even), leading coefficient $4$ (positive). Both ends go up: $g(x) \to +\infty$ as $x \to \pm\infty$.
(c) $h(x) = -3(x-1)^2(x+2)^3$
Expanding the leading term: $-3 \cdot x^2 \cdot x^3 = -3x^5$. Same as case (a): left end up, right end down.
1.2 Remainder and Factor Theorems
Remainder Theorem & Factor Theorem
Remainder Theorem: When $f(x)$ is divided by $(x - a)$, the remainder equals $f(a)$. You can find the remainder without doing the division.
Factor Theorem: $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$.
Rational Root Theorem: If $f(x) = a_n x^n + \cdots + a_0$ has integer coefficients, any rational zero $\frac{p}{q}$ (in lowest terms) satisfies: $p \mid a_0$ and $q \mid a_n$.
Worked Example 1.2 — Using the Remainder Theorem
Find the remainder when $f(x) = 3x^4 - 2x^3 + x - 5$ is divided by $(x + 2)$.
By the Remainder Theorem, substitute $x = -2$: $$f(-2) = 3(-2)^4 - 2(-2)^3 + (-2) - 5 = 3(16) - 2(-8) - 2 - 5 = 48 + 16 - 2 - 5 = 57$$
The remainder is $\mathbf{57}$. (Notice we never performed long division.)
1.3 Factoring Polynomials: The Step-by-Step Process
The standard approach: (1) list possible rational roots using $\pm\frac{p}{q}$; (2) test each with the Factor Theorem; (3) once a root $a$ is found, divide out $(x - a)$ using synthetic division; (4) repeat on the quotient.
Worked Example 1.3 — Factoring a Cubic Completely
Factor $f(x) = x^3 - 2x^2 - 5x + 6$ completely and state all real zeros.
Step 1 Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6$ (factors of $6$, leading coeff $= 1$).
Step 2 Test: $f(1) = 1 - 2 - 5 + 6 = 0$. $(x - 1)$ is a factor.
Step 3 Synthetic division by $(x - 1)$:
1 | 1 −2 −5 6
| 1 −1 −6
| 1 −1 −6 0
Quotient: $x^2 - x - 6$.
Step 4 Factor the quadratic: $x^2 - x - 6 = (x - 3)(x + 2)$.
$$\boxed{f(x) = (x-1)(x-3)(x+2)}$$
Zeros: $x = 1, 3, -2$. All have multiplicity 1, so the graph crosses the $x$-axis at each zero.
Worked Example 1.4 — Factoring a Quartic
Factor $f(x) = x^4 - x^3 - 7x^2 + x + 6$ completely.
Step 1 Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6$.
Step 2 Test $x = 1$: $f(1) = 1 - 1 - 7 + 1 + 6 = 0$. So $(x-1)$ is a factor.
Step 3 Synthetic division by $1$: quotient is $x^3 - 7x - 6$.
Step 4 Factor $g(x) = x^3 - 7x - 6$. Test $x = -1$: $g(-1) = -1 + 7 - 6 = 0$. So $(x + 1)$ is a factor.
Step 5 Divide $x^3 - 7x - 6$ by $(x+1)$: quotient is $x^2 - x - 6 = (x-3)(x+2)$.
$$\boxed{f(x) = (x-1)(x+1)(x-3)(x+2)}$$
Zeros: $x = 1, -1, 3, -2$. Degree 4 with positive leading coefficient: both ends up.
1.4 Multiplicity and Graph Behaviour
Multiplicity Rules
- Odd multiplicity (1, 3, 5, …): graph crosses the $x$-axis at the zero.
- Even multiplicity (2, 4, 6, …): graph touches the $x$-axis and turns back (bounces).
- The sum of all multiplicities equals the degree of the polynomial.
Worked Example 1.5 — Sketch from Factored Form
Sketch $f(x) = -2(x + 3)(x - 1)^2(x - 4)$ and solve $f(x) \geq 0$.
Step 1 — Zeros & multiplicity $x = -3$ (mult 1, crosses); $x = 1$ (mult 2, bounces); $x = 4$ (mult 1, crosses).
Step 2 — End behaviour Leading term: $-2 \cdot x \cdot x^2 \cdot x = -2x^4$. Degree 4, negative: both ends down.
Step 3 — $y$-intercept $f(0) = -2(3)(1)(-4) = 24$.
Step 4 — Sign chart Critical values: $-3, 1, 4$.
| Interval | $x+3$ | $(x-1)^2$ | $x-4$ | $-2(\cdots)$ | $f(x)$ |
|---|---|---|---|---|---|
| $x < -3$ | − | + | − | $-2(-)(+)(-) = -$ | neg |
| $-3 < x < 1$ | + | + | − | $-2(+)(+)(-) = +$ | pos |
| $1 < x < 4$ | + | + | − | $-2(+)(+)(-) = +$ | pos |
| $x > 4$ | + | + | + | $-2(+)(+)(+) = -$ | neg |
Step 5 — Solution $f(x) \geq 0$ (including zeros): $x \in [-3,\, 1] \cup [1,\, 4] = [-3,\, 4]$.
Note Although $x = 1$ is included in the solution, it is a bouncing zero — the function merely touches zero there before rising back up.
2. Rational Functions
A rational function is a ratio of two polynomials: $f(x) = \dfrac{p(x)}{q(x)}$, $q(x) \neq 0$. The asymptotic and discontinuity behaviour all come from the interaction between $p$ and $q$.
Asymptote Reference Card
- Vertical asymptote (VA): $x = a$ where $q(a) = 0$ and $p(a) \neq 0$. If both are zero, factor and cancel — you get a hole instead.
- Horizontal asymptote (HA):
- $\deg p < \deg q$: $\quad y = 0$
- $\deg p = \deg q$: $\quad y = \dfrac{\text{leading coeff of } p}{\text{leading coeff of } q}$
- $\deg p > \deg q$: $\quad$ no HA (oblique or polynomial asymptote)
- Oblique (slant) asymptote: exists when $\deg p = \deg q + 1$; found by polynomial long division. The remainder $\to 0$ as $x \to \pm\infty$.
Worked Example 2.1 — Identifying Asymptotes (No Cancellation)
Find all asymptotes, intercepts, and the domain of $f(x) = \dfrac{3x + 6}{x^2 - x - 2}$.
Step 1 — Factor both Numerator: $3(x+2)$. Denominator: $(x-2)(x+1)$.
Step 2 — Check for holes No common factors $\Rightarrow$ no holes.
Step 3 — VAs Denominator zero at $x = 2$ and $x = -1$ (numerator $\neq 0$ at both). VAs: $x = 2$ and $x = -1$.
Step 4 — HA $\deg p = 1 < \deg q = 2$: HA is $y = 0$.
Step 5 — Intercepts $x$-intercept: set numerator $= 0 \Rightarrow x = -2$, so $(-2,0)$. $y$-intercept: $f(0) = \dfrac{6}{-2} = -3$, so $(0,-3)$.
Domain: $\{x \in \mathbb{R} \mid x \neq 2, x \neq -1\}$.
Worked Example 2.2 — Hole vs. Vertical Asymptote
Analyze $f(x) = \dfrac{x^2 - x - 6}{x^2 - 4}$ completely.
Step 1 — Factor $\dfrac{(x-3)(x+2)}{(x-2)(x+2)}$. Common factor: $(x+2)$.
Step 2 — Hole Cancel $(x+2)$: simplified form is $\dfrac{x-3}{x-2}$ for $x \neq -2$. There is a hole at $x = -2$. $y$-coordinate: $\dfrac{-2-3}{-2-2} = \dfrac{-5}{-4} = \dfrac{5}{4}$. Hole at $\left(-2,\tfrac{5}{4}\right)$.
Step 3 — VA $x = 2$ (denominator of simplified form).
Step 4 — HA Equal degrees after cancellation; leading coefficients both $1$: HA is $y = 1$.
Step 5 — Intercepts $x$-intercept: numerator $= 0 \Rightarrow x = 3$, so $(3, 0)$. $y$-intercept: $f(0) = \dfrac{-3}{-2} = \dfrac{3}{2}$, so $\left(0, \tfrac{3}{2}\right)$.
Worked Example 2.3 — Oblique Asymptote
Find all asymptotes of $g(x) = \dfrac{2x^2 - 5x + 1}{x - 3}$.
Step 1 — Check degrees $\deg p = 2 = \deg q + 1 = 1 + 1$. There is an oblique asymptote.
Step 2 — Long division $$2x^2 - 5x + 1 \div (x - 3)$$ $$2x^2 - 5x + 1 = (x-3)(2x + 1) + 4$$ So $g(x) = 2x + 1 + \dfrac{4}{x-3}$.
Step 3 As $x \to \pm\infty$, $\dfrac{4}{x-3} \to 0$. Oblique asymptote: $y = 2x + 1$.
Step 4 — VA $x = 3$ (denominator zero, no cancellation).
Worked Example 2.4 — Rational Inequality via Sign Chart
Solve $\dfrac{(x+1)(x-3)}{(x-1)(x+2)} \geq 0$.
Step 1 Critical values (zeros of numerator and denominator): $x = -2, -1, 1, 3$. Note: $x = -2$ and $x = 1$ are excluded from the domain.
Step 2 — Sign chart
| Interval | $x+2$ | $x+1$ | $x-1$ | $x-3$ | Result |
|---|---|---|---|---|---|
| $x < -2$ | − | − | − | − | + |
| $-2 < x < -1$ | + | − | − | − | − |
| $-1 < x < 1$ | + | + | − | − | + |
| $1 < x < 3$ | + | + | + | − | − |
| $x > 3$ | + | + | + | + | + |
Step 3 Include zeros of numerator ($x = -1$, $x = 3$); exclude zeros of denominator ($x = -2$, $x = 1$).
$$\boxed{x \in (-\infty,\,-2) \cup [-1,\,1) \cup [3,\,+\infty)}$$
3. Exponential and Logarithmic Functions
Exponential and logarithmic functions are inverses of each other. The key to solving equations in this family is recognizing which "direction" to convert: rewrite with the same base (if possible) or take logarithms of both sides.
Definition and Laws
$\log_b x = y \iff b^y = x$ (for $b > 0$, $b \neq 1$, $x > 0$).
- Product: $\log_b(MN) = \log_b M + \log_b N$
- Quotient: $\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N$
- Power: $\log_b(M^p) = p\log_b M$
- Change of base: $\log_b M = \dfrac{\log M}{\log b} = \dfrac{\ln M}{\ln b}$
Domain warning: Logarithms require positive arguments. Always check your solutions by substituting back — extraneous solutions arise whenever you square both sides or combine logarithms.
Worked Example 3.1 — Same Base: Rewrite and Equate Exponents
Solve $8^{x-1} = 4^{2x-3}$.
Step 1 Rewrite both sides as powers of $2$: $8 = 2^3$ and $4 = 2^2$.
$$2^{3(x-1)} = 2^{2(2x-3)}$$
Step 2 Equate exponents: $$3(x-1) = 2(2x-3) \implies 3x - 3 = 4x - 6 \implies x = 3$$
Check: $8^{3-1} = 8^2 = 64$ and $4^{2(3)-3} = 4^3 = 64$. ✓
Worked Example 3.2 — Substitution Method
Solve $9^x - 4 \cdot 3^x + 3 = 0$.
Key observation $9^x = (3^2)^x = (3^x)^2$. Let $u = 3^x$ (note $u > 0$ always).
Step 1 Substitute: $u^2 - 4u + 3 = 0$.
Step 2 Factor: $(u - 1)(u - 3) = 0$, so $u = 1$ or $u = 3$.
Step 3 Back-substitute:
- $3^x = 1 = 3^0 \Rightarrow x = 0$
- $3^x = 3 = 3^1 \Rightarrow x = 1$
Solutions: $x = 0$ and $x = 1$.
Worked Example 3.3 — Different Bases: Take Logarithms
Solve $3^{2x-1} = 5^{x+2}$. Give an exact answer and a decimal approximation.
Step 1 Take $\log$ of both sides: $$(2x - 1)\log 3 = (x + 2)\log 5$$
Step 2 Expand and collect $x$-terms: $$2x\log 3 - \log 3 = x\log 5 + 2\log 5$$ $$x(2\log 3 - \log 5) = 2\log 5 + \log 3$$
Step 3 Solve: $$x = \frac{2\log 5 + \log 3}{2\log 3 - \log 5} = \frac{\log 75}{\log(9/5)} \approx \frac{1.8751}{0.2553} \approx 7.35$$
Worked Example 3.4 — Logarithmic Equation (Single Log)
Solve $2\log_3(x - 1) = \log_3(x + 5)$.
Step 1 Apply the power law to the left side: $\log_3(x-1)^2 = \log_3(x+5)$.
Step 2 Equate arguments (same base, both sides are logarithms): $$(x-1)^2 = x + 5 \implies x^2 - 2x + 1 = x + 5 \implies x^2 - 3x - 4 = 0$$
Step 3 Factor: $(x - 4)(x + 1) = 0$, so $x = 4$ or $x = -1$.
Step 4 — Domain check
- $x = 4$: $\log_3(3) = 1$ ✓ and $\log_3(9) = 2$ ✓. Valid.
- $x = -1$: $\log_3(-2)$ is undefined. Reject $x = -1$.
Answer: $x = 4$.
Worked Example 3.5 — Multiple Logs Leading to a Quadratic
Solve $\log(x + 2) + \log(x - 1) = 1$.
Step 1 Product law: $\log[(x+2)(x-1)] = 1$.
Step 2 Convert to exponential form ($\log = \log_{10}$): $(x+2)(x-1) = 10$.
Step 3 Expand: $x^2 + x - 2 = 10 \implies x^2 + x - 12 = 0 \implies (x+4)(x-3) = 0$.
Step 4 — Domain check Both arguments must be positive:
- $x = 3$: $\log(5)$ and $\log(2)$ — both valid. ✓
- $x = -4$: $\log(-2)$ — undefined. Reject.
Answer: $x = 3$.
4. Trigonometric Functions and Identities
MHF4U pushes beyond MCR3U by introducing reciprocal trig functions, the full toolkit of identities (Pythagorean, compound angle, double angle), and the ability to both prove identities and solve equations using them. The worked examples build from reading exact values up to multi-step proofs.
Identity Reference Sheet
Reciprocal: $\csc\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$
Pythagorean:
- $\sin^2\theta + \cos^2\theta = 1$
- $\tan^2\theta + 1 = \sec^2\theta$
- $\cot^2\theta + 1 = \csc^2\theta$
Compound Angle:
- $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$
- $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$
- $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$
Double Angle:
- $\sin 2A = 2\sin A\cos A$
- $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
Worked Example 4.1 — Exact Values with Special Angles
Evaluate $\cos\dfrac{5\pi}{6}$ and $\tan\dfrac{7\pi}{4}$ exactly without a calculator.
$\cos\dfrac{5\pi}{6}$: Reference angle $= \pi - \dfrac{5\pi}{6} = \dfrac{\pi}{6}$. In Q2, cosine is negative. $\cos\dfrac{5\pi}{6} = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}$.
$\tan\dfrac{7\pi}{4}$: Reference angle $= 2\pi - \dfrac{7\pi}{4} = \dfrac{\pi}{4}$. In Q4, tangent is negative. $\tan\dfrac{7\pi}{4} = -\tan\dfrac{\pi}{4} = -1$.
Worked Example 4.2 — Exact Value Using Compound Angle
Find the exact value of $\sin 75°$.
Write $75° = 45° + 30°$ and apply the sine addition formula: $$\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$
Alternative $75° = 45° + 30°$ or $= 120° - 45°$. Either decomposition works.
Worked Example 4.3 — Double Angle Given One Trig Value
Given $\sin\theta = \dfrac{3}{5}$ with $0 < \theta < \dfrac{\pi}{2}$, find $\sin 2\theta$, $\cos 2\theta$, and $\tan 2\theta$.
Step 1 — Find $\cos\theta$ Using $\sin^2\theta + \cos^2\theta = 1$: $$\cos\theta = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ (positive because $\theta$ is in Q1).
Step 2 $$\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$$ $$\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \cdot \frac{9}{25} = 1 - \frac{18}{25} = \frac{7}{25}$$ $$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{24/25}{7/25} = \frac{24}{7}$$
Worked Example 4.4 — Proving a Trigonometric Identity
Prove: $\dfrac{\cos 2\theta}{1 + \sin 2\theta} = \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$.
Strategy Always work on one side only. Work on the LHS, expanding using double-angle formulas.
Apply $\cos 2\theta = \cos^2\theta - \sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$: $$\text{LHS} = \frac{\cos^2\theta - \sin^2\theta}{1 + 2\sin\theta\cos\theta}$$
Factor numerator as difference of squares, and recognize denominator as a perfect square: $$= \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{(\cos\theta + \sin\theta)^2}$$
Cancel the common factor $(\cos\theta + \sin\theta)$: $$= \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \text{RHS} \quad\checkmark$$
Worked Example 4.5 — Solving a Trig Equation (Quadratic Form)
Solve $2\sin^2 x - \sin x - 1 = 0$ for $x \in [0, 2\pi]$.
Step 1 — Substitute Let $u = \sin x$: $2u^2 - u - 1 = 0$.
Step 2 — Factor $(2u + 1)(u - 1) = 0 \Rightarrow u = -\dfrac{1}{2}$ or $u = 1$.
Step 3 — Find $x$
- $\sin x = 1$: $x = \dfrac{\pi}{2}$.
- $\sin x = -\dfrac{1}{2}$: reference angle $= \dfrac{\pi}{6}$; sine negative in Q3 and Q4: $x = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$ and $x = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$.
Solutions: $x \in \left\{\dfrac{\pi}{2},\, \dfrac{7\pi}{6},\, \dfrac{11\pi}{6}\right\}$.
5. Characteristics of Functions: Composition and Inverses
Composition of Functions
$(f \circ g)(x) = f(g(x))$: apply $g$ first, then $f$ to the output.
Domain of $f \circ g$: the set of all $x$ in Dom$(g)$ such that $g(x) \in$ Dom$(f)$. This requires a two-step check.
Key fact: $f \circ g \neq g \circ f$ in general (composition is not commutative).
Worked Example 5.1 — Basic Composition and Domain
Let $f(x) = x^2 + 1$ and $g(x) = \sqrt{x - 2}$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$ with domains.
$(f \circ g)(x) = f(\sqrt{x-2}) = (\sqrt{x-2})^2 + 1 = x - 2 + 1 = x - 1$
Domain: need $x - 2 \geq 0$, so $x \geq 2$. Domain: $[2, +\infty)$.
$(g \circ f)(x) = g(x^2 + 1) = \sqrt{(x^2 + 1) - 2} = \sqrt{x^2 - 1}$
Domain: need $x^2 - 1 \geq 0$, so $|x| \geq 1$. Domain: $(-\infty, -1] \cup [1, +\infty)$.
Confirm: $f \circ g \neq g \circ f$.
Worked Example 5.2 — Careful Domain Restriction
Let $f(x) = \dfrac{1}{x}$ (defined for $x \neq 0$) and $g(x) = x - 4$ (defined for all $x$).
Find $(f \circ g)(x)$ and state its domain carefully.
$(f \circ g)(x) = f(x - 4) = \dfrac{1}{x - 4}$.
Domain: start with all $x$; then require $g(x) = x - 4 \in$ Dom$(f)$, i.e., $x - 4 \neq 0$, i.e., $x \neq 4$.
Domain of $f \circ g$: $\{x \in \mathbb{R} \mid x \neq 4\}$.
Now find $(g \circ f)(x) = g\!\left(\dfrac{1}{x}\right) = \dfrac{1}{x} - 4$.
Domain: require $x \in$ Dom$(f)$, i.e., $x \neq 0$. Domain: $\{x \in \mathbb{R} \mid x \neq 0\}$.
Inverse Functions
$f^{-1}$ exists if and only if $f$ is one-to-one (passes the horizontal line test). To find $f^{-1}$:
- Replace $f(x)$ with $y$.
- Swap $x$ and $y$.
- Solve for $y$ — this is $f^{-1}(x)$.
- Domain of $f^{-1}$ = Range of $f$; Range of $f^{-1}$ = Domain of $f$.
Verification: $(f^{-1} \circ f)(x) = x$ for all $x$ in Dom$(f)$, and $(f \circ f^{-1})(x) = x$ for all $x$ in Dom$(f^{-1})$.
Worked Example 5.3 — Finding an Inverse with Domain Restriction
Find the inverse of $f(x) = x^2 - 4$ restricted to $x \geq 0$. State the domain and range of $f^{-1}$.
Step 1 With restriction $x \geq 0$: Range of $f$ is $[-4, +\infty)$.
Step 2 Swap $x$ and $y$ in $y = x^2 - 4$: $x = y^2 - 4$.
Step 3 Solve for $y$: $y^2 = x + 4$, so $y = \sqrt{x + 4}$ (positive root, since original restriction was $x \geq 0$).
$$f^{-1}(x) = \sqrt{x + 4}$$
Domain of $f^{-1}$: $[-4, +\infty)$ | Range of $f^{-1}$: $[0, +\infty)$.
Worked Example 5.4 — Verifying Inverses
Show that $f(x) = 2x + 5$ and $g(x) = \dfrac{x - 5}{2}$ are inverses of each other.
$(f \circ g)(x)$: $$f(g(x)) = f\!\left(\frac{x-5}{2}\right) = 2 \cdot \frac{x-5}{2} + 5 = (x - 5) + 5 = x \quad\checkmark$$
$(g \circ f)(x)$: $$g(f(x)) = g(2x + 5) = \frac{(2x+5) - 5}{2} = \frac{2x}{2} = x \quad\checkmark$$
Both compositions return $x$, confirming $g = f^{-1}$.
6. Practice Problems
Forty problems organized by topic. Attempt each problem before revealing the solution. Within each group, problems progress from foundational to exam-level.
Problem 1 — Remainder Theorem
Without performing division, find the remainder when $f(x) = x^4 - 3x^3 + 2x - 5$ is divided by $(x - 2)$.
Show Solution
$f(2) = 16 - 24 + 4 - 5 = -9$. The remainder is $\mathbf{-9}$.
Problem 2 — Factor Theorem
Determine whether $(x + 3)$ is a factor of $f(x) = 2x^3 + 5x^2 - x - 6$. If yes, find all zeros.
Show Solution
$f(-3) = 2(-27) + 5(9) - (-3) - 6 = -54 + 45 + 3 - 6 = -12 \neq 0$. Not a factor.
Try $x = 1$: $f(1) = 2 + 5 - 1 - 6 = 0$. So $(x-1)$ is a factor.
Divide: $f(x) = (x-1)(2x^2 + 7x + 6) = (x-1)(2x+3)(x+2)$.
Zeros: $x = 1,\; x = -\dfrac{3}{2},\; x = -2$.
Problem 3 — Factoring a Cubic
Factor $f(x) = 2x^3 + x^2 - 13x + 6$ completely and state all real zeros.
Show Solution
Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2}$.
$f(2) = 16 + 4 - 26 + 6 = 0$. Factor out $(x - 2)$.
Synthetic division: quotient is $2x^2 + 5x - 3$.
Factor: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$.
$f(x) = (x-2)(2x-1)(x+3)$. Zeros: $x = 2,\; \dfrac{1}{2},\; -3$.
Problem 4 — Determining Equation from Zeros
A polynomial of degree 4 has zeros at $x = -1$ (multiplicity 2), $x = 3$, and $x = 5$, and passes through $(0, -30)$. Find the equation.
Show Solution
General form: $f(x) = a(x+1)^2(x-3)(x-5)$.
Use $(0, -30)$: $-30 = a(1)^2(-3)(-5) = 15a$, so $a = -2$.
$f(x) = -2(x+1)^2(x-3)(x-5)$.
Problem 5 — End Behaviour
Describe the end behaviour of $f(x) = -(x-1)^2(x+3)^3$ and find the $y$-intercept.
Show Solution
Leading term: $-(x^2)(x^3) = -x^5$. Degree 5 (odd), negative leading coeff: as $x\to+\infty$, $f\to-\infty$; as $x\to-\infty$, $f\to+\infty$.
$y$-intercept: $f(0) = -(0-1)^2(0+3)^3 = -(1)(27) = -27$. So $(0, -27)$.
Problem 6 — Multiplicity and Graph Behaviour
For $f(x) = (x+2)^2(x-1)(x-4)^3$, state the degree, end behaviour, and whether the graph crosses or bounces at each zero.
Show Solution
Degree: $2 + 1 + 3 = 6$. Leading coeff: $+1$ (positive). Both ends up.
$x = -2$ (mult 2): bounces. $x = 1$ (mult 1): crosses. $x = 4$ (mult 3): crosses (odd multiplicity, though flattened).
Problem 7 — Polynomial Inequality
Solve $(x - 2)(x + 1)(x - 5) \leq 0$.
Show Solution
Zeros: $x = -1, 2, 5$ (all multiplicity 1). Leading term: $x^3$ (positive, odd) — right end up.
| Interval | Sign of product |
|---|---|
| $x < -1$ | negative |
| $-1 < x < 2$ | positive |
| $2 < x < 5$ | negative |
| $x > 5$ | positive |
Solution ($\leq 0$): $x \in (-\infty,\, -1] \cup [2,\, 5]$.
Problem 8 — Finding a Value via the Remainder Theorem
Given that $f(x) = x^3 + kx^2 - 2x + 1$ has remainder $13$ when divided by $(x - 2)$, find the value of $k$.
Show Solution
By the Remainder Theorem: $f(2) = 13$.
$8 + 4k - 4 + 1 = 13 \implies 4k + 5 = 13 \implies 4k = 8 \implies k = 2$.
Problem 9 — Identifying Asymptotes
Find all asymptotes of $f(x) = \dfrac{2x + 4}{x - 3}$.
Show Solution
VA: $x = 3$ (denominator zero, numerator $2(3)+4=10 \neq 0$).
HA: equal degrees; ratio of leading coefficients $= \dfrac{2}{1} = 2$. HA: $y = 2$.
Problem 10 — Hole vs. Asymptote
Find all asymptotes and holes of $f(x) = \dfrac{x^2 - 9}{x^2 + x - 6}$.
Show Solution
Factor: $\dfrac{(x-3)(x+3)}{(x+3)(x-2)}$. Common factor $(x+3)$ → hole at $x = -3$.
$y$-coordinate of hole: $\dfrac{-3-3}{-3-2} = \dfrac{-6}{-5} = \dfrac{6}{5}$. Hole: $\left(-3,\,\dfrac{6}{5}\right)$.
Simplified: $\dfrac{x-3}{x-2}$. VA: $x = 2$. HA: $y = 1$.
Problem 11 — $y$-intercept and $x$-intercept
Find all intercepts of $f(x) = \dfrac{x^2 - 4x + 3}{x^2 - 1}$.
Show Solution
Factor: $\dfrac{(x-1)(x-3)}{(x-1)(x+1)}$. Hole at $x = 1$; simplified: $\dfrac{x-3}{x+1}$.
$y$-intercept: $f(0) = \dfrac{-3}{1} = -3$. So $(0,-3)$.
$x$-intercept: numerator of simplified $= 0 \Rightarrow x = 3$. So $(3, 0)$.
VA: $x = -1$. HA: $y = 1$.
Problem 12 — Oblique Asymptote
Find the oblique asymptote of $h(x) = \dfrac{x^2 - x + 2}{x + 2}$.
Show Solution
Polynomial division: $x^2 - x + 2 = (x+2)(x - 3) + 8$.
So $h(x) = x - 3 + \dfrac{8}{x+2}$. As $x \to \pm\infty$, the fraction vanishes.
Oblique asymptote: $y = x - 3$. VA: $x = -2$.
Problem 13 — Rational Inequality (Basic)
Solve $\dfrac{x - 2}{x + 3} \geq 1$.
Show Solution
Rearrange: $\dfrac{x-2}{x+3} - 1 \geq 0 \Rightarrow \dfrac{x-2-(x+3)}{x+3} \geq 0 \Rightarrow \dfrac{-5}{x+3} \geq 0$.
Since $-5 < 0$, need $x + 3 < 0$, i.e., $x < -3$.
Solution: $x \in (-\infty,\, -3)$.
Problem 14 — Rational Inequality (Sign Chart)
Solve $\dfrac{x^2 - 4}{x - 1} < 0$.
Show Solution
Factor numerator: $\dfrac{(x-2)(x+2)}{x-1}$. Critical values: $-2, 1, 2$ (VA at $x=1$).
| Interval | $(x+2)$ | $(x-2)$ | $(x-1)$ | Result |
|---|---|---|---|---|
| $x < -2$ | − | − | − | − |
| $-2 < x < 1$ | + | − | − | + |
| $1 < x < 2$ | + | − | + | − |
| $x > 2$ | + | + | + | + |
Solution ($< 0$): $x \in (-\infty,\,-2) \cup (1,\,2)$.
Problem 15 — Domain of a Rational Function
State the domain of $f(x) = \dfrac{\sqrt{x + 5}}{x^2 - 4}$.
Show Solution
Numerator requires $x + 5 \geq 0 \Rightarrow x \geq -5$.
Denominator requires $x^2 - 4 \neq 0 \Rightarrow x \neq \pm 2$.
Domain: $[-5,\, +\infty) \setminus \{-2,\, 2\} = [-5,\,-2) \cup (-2,\,2) \cup (2,\,+\infty)$.
Problem 16 — Constructing a Rational Function
Write a rational function $f(x)$ that has: VA at $x = 1$ and $x = -3$; a hole at $x = 2$; HA at $y = 0$; $x$-intercept at $x = -1$.
Show Solution
HA at $y = 0$ means $\deg p < \deg q$. Hole at $x = 2$ means $(x-2)$ is a common factor. VAs at $x=1$ and $x=-3$ mean $(x-1)(x+3)$ is in the denominator after cancellation. $x$-intercept at $x=-1$ means $(x+1)$ is in the numerator.
One answer: $f(x) = \dfrac{(x+1)(x-2)}{(x-2)(x-1)(x+3)}$.
Check degrees: numerator deg 2, denominator deg 3. HA: $y=0$ ✓.
Problem 17 — Same-Base Exponential
Solve $27^{x+1} = 9^{3x-1}$.
Show Solution
$3^{3(x+1)} = 3^{2(3x-1)} \Rightarrow 3x + 3 = 6x - 2 \Rightarrow 5 = 3x \Rightarrow x = \dfrac{5}{3}$.
Problem 18 — Substitution (Exponential Quadratic)
Solve $4^x - 6 \cdot 2^x + 8 = 0$.
Show Solution
$4^x = (2^x)^2$. Let $u = 2^x$: $u^2 - 6u + 8 = 0 \Rightarrow (u-2)(u-4) = 0$.
$2^x = 2 \Rightarrow x = 1$; $2^x = 4 \Rightarrow x = 2$.
Problem 19 — Different Bases
Solve $5^{2x} = 7^{x-1}$. Give an exact answer using logarithms.
Show Solution
$2x \log 5 = (x-1)\log 7 \Rightarrow 2x\log 5 = x\log 7 - \log 7$.
$x(2\log 5 - \log 7) = -\log 7 \Rightarrow x = \dfrac{-\log 7}{2\log 5 - \log 7} = \dfrac{\log 7}{\log 7 - \log 25} = \dfrac{\log 7}{\log(7/25)}$.
Problem 20 — Log Laws (Simplification)
Simplify $\log_4 8 + \log_4 2 - \log_4 \sqrt{2}$ to a single rational number.
Show Solution
$\log_4 8 + \log_4 2 - \log_4 2^{1/2} = \log_4\!\left(\dfrac{8 \cdot 2}{2^{1/2}}\right) = \log_4\!\left(\dfrac{16}{\sqrt{2}}\right) = \log_4\!\left(2^{7/2}\right)$.
$= \dfrac{7}{2}\log_4 2 = \dfrac{7}{2} \cdot \dfrac{1}{2} = \dfrac{7}{4}$.
Problem 21 — Log Equation (Single Log)
Solve $\log_5(3x - 1) = 2$.
Show Solution
Convert: $3x - 1 = 5^2 = 25 \Rightarrow 3x = 26 \Rightarrow x = \dfrac{26}{3}$.
Check: $3 \cdot \dfrac{26}{3} - 1 = 25 > 0$ ✓.
Problem 22 — Log Equation (Multiple Logs)
Solve $\log(x+2) + \log(x-1) = 1$.
Show Solution
$\log[(x+2)(x-1)] = 1 \Rightarrow (x+2)(x-1) = 10 \Rightarrow x^2 + x - 2 = 10$.
$x^2 + x - 12 = 0 \Rightarrow (x+4)(x-3) = 0$.
$x = -4$: gives $\log(-2)$ — invalid. $x = 3$: valid ✓. Answer: $x = 3$.
Problem 23 — Log Equation (Quadratic After Combining)
Solve $2\log_3(x-1) = \log_3(x+5)$.
Show Solution
$\log_3(x-1)^2 = \log_3(x+5) \Rightarrow (x-1)^2 = x+5$.
$x^2 - 3x - 4 = 0 \Rightarrow (x-4)(x+1) = 0$.
$x = -1$: $\log_3(-2)$ undefined. Reject. $x = 4$: valid ✓.
Problem 24 — Change of Base
Evaluate $\log_8 125$ using the change of base formula. Leave as an exact simplified fraction if possible; otherwise give a 4-decimal approximation.
Show Solution
$\log_8 125 = \dfrac{\log 125}{\log 8} = \dfrac{3\log 5}{3\log 2} = \dfrac{\log 5}{\log 2} = \log_2 5 \approx \dfrac{0.6990}{0.3010} \approx 2.3219$.
Problem 25 — Log Inequality
Solve $\log_3(x - 2) < 2$. State the solution as an interval and identify any domain restrictions.
Show Solution
Domain: $x - 2 > 0 \Rightarrow x > 2$.
Since $\log_3$ is increasing: $x - 2 < 3^2 = 9 \Rightarrow x < 11$.
Combining: $2 < x < 11$. Solution: $(2,\, 11)$.
Problem 26 — Applied: Half-Life
A radioactive substance decays according to $A(t) = A_0 \cdot 2^{-t/k}$, where $t$ is in years and $k$ is the half-life. If 80 g decays to 5 g in 24 years, find the half-life $k$, exact.
Show Solution
$5 = 80 \cdot 2^{-24/k} \Rightarrow \dfrac{1}{16} = 2^{-24/k} \Rightarrow 2^{-4} = 2^{-24/k} \Rightarrow -4 = -\dfrac{24}{k} \Rightarrow k = 6$.
The half-life is $\mathbf{6}$ years.
Problem 27 — Exact Values
Without a calculator, evaluate $\sin\dfrac{5\pi}{3}$ and $\cos\dfrac{3\pi}{4}$.
Show Solution
$\sin\dfrac{5\pi}{3}$: reference angle $\dfrac{\pi}{3}$, Q4, sine negative: $-\dfrac{\sqrt{3}}{2}$.
$\cos\dfrac{3\pi}{4}$: reference angle $\dfrac{\pi}{4}$, Q2, cosine negative: $-\dfrac{\sqrt{2}}{2}$.
Problem 28 — Compound Angle (Exact Value)
Find the exact value of $\cos 15°$.
Show Solution
Write $15° = 45° - 30°$: $$\cos 15° = \cos 45°\cos 30° + \sin 45°\sin 30° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$$
Problem 29 — Double Angle from Given Value
Given $\cos\theta = -\dfrac{5}{13}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find $\sin 2\theta$ and $\cos 2\theta$.
Show Solution
$\theta$ in Q3: $\sin\theta = -\sqrt{1 - \frac{25}{169}} = -\dfrac{12}{13}$.
$\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\left(-\dfrac{12}{13}\right)\cdot\left(-\dfrac{5}{13}\right) = \dfrac{120}{169}$.
$\cos 2\theta = 2\cos^2\theta - 1 = 2\cdot\dfrac{25}{169} - 1 = \dfrac{50}{169} - 1 = -\dfrac{119}{169}$.
Problem 30 — Prove a Pythagorean Identity
Prove: $\sec^2\theta - \tan^2\theta \cdot \sec^2\theta = 1 - \tan^2\theta$.
Show Solution
LHS $= \sec^2\theta(1 - \tan^2\theta)$. Use $\sec^2\theta = 1 + \tan^2\theta$: $= (1 + \tan^2\theta)(1 - \tan^2\theta) = 1 - \tan^4\theta$.
Hmm — this doesn't simplify to $1 - \tan^2\theta$ directly. Let's work on the RHS instead to match LHS.
Actually, factor the LHS differently: LHS $= \sec^2\theta - \tan^2\theta\sec^2\theta = \sec^2\theta(1 - \tan^2\theta)$. RHS $= 1 - \tan^2\theta$.
For these to be equal: $\sec^2\theta(1-\tan^2\theta) = 1 - \tan^2\theta$ requires $\sec^2\theta = 1$ or $\tan\theta = \pm 1$ — so this is not an identity. Note for students: Check the problem statement carefully before attempting to prove. This is an example of a statement that is conditional, not an identity.
Problem 31 — Proving an Identity (Compound Angle)
Prove: $\dfrac{\sin(A+B)}{\cos A \cos B} = \tan A + \tan B$.
Show Solution
LHS $= \dfrac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \dfrac{\sin A \cos B}{\cos A \cos B} + \dfrac{\cos A \sin B}{\cos A \cos B} = \dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B} = \tan A + \tan B$ = RHS. $\checkmark$
Problem 32 — Solving a Basic Trig Equation
Solve $\cos 2x = \cos x$ for $x \in [0, 2\pi]$.
Show Solution
Use $\cos 2x = 2\cos^2 x - 1$: $2\cos^2 x - 1 = \cos x \Rightarrow 2\cos^2 x - \cos x - 1 = 0$.
$(2\cos x + 1)(\cos x - 1) = 0$. So $\cos x = 1$ or $\cos x = -\dfrac{1}{2}$.
$\cos x = 1 \Rightarrow x = 0,\; 2\pi$. $\cos x = -\dfrac{1}{2} \Rightarrow x = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$.
Solutions: $x \in \left\{0,\;\dfrac{2\pi}{3},\;\dfrac{4\pi}{3},\;2\pi\right\}$.
Problem 33 — Trig Equation (Quadratic, Multiple Approaches)
Solve $\tan^2 x - 3 = 0$ for $x \in [0, 2\pi]$.
Show Solution
$\tan^2 x = 3 \Rightarrow \tan x = \pm\sqrt{3}$.
$\tan x = \sqrt{3}$: reference angle $\dfrac{\pi}{3}$, positive in Q1 and Q3: $x = \dfrac{\pi}{3},\;\dfrac{4\pi}{3}$.
$\tan x = -\sqrt{3}$: negative in Q2 and Q4: $x = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$, and $x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$.
Solutions: $x \in \left\{\dfrac{\pi}{3},\;\dfrac{2\pi}{3},\;\dfrac{4\pi}{3},\;\dfrac{5\pi}{3}\right\}$.
Problem 34 — Proving a Double Angle Identity
Prove: $\dfrac{\sin 2\theta}{\sin\theta} - \dfrac{\cos 2\theta}{\cos\theta} = \sec\theta$.
Show Solution
LHS $= \dfrac{2\sin\theta\cos\theta}{\sin\theta} - \dfrac{\cos^2\theta - \sin^2\theta}{\cos\theta} = 2\cos\theta - \dfrac{\cos^2\theta - \sin^2\theta}{\cos\theta}$.
$= \dfrac{2\cos^2\theta - (\cos^2\theta - \sin^2\theta)}{\cos\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\cos\theta} = \dfrac{1}{\cos\theta} = \sec\theta$ = RHS. $\checkmark$
Problem 35 — Evaluating a Composition
Let $f(x) = 2x - 1$ and $g(x) = x^2 + 3$. Evaluate $(f \circ g)(2)$ and $(g \circ f)(-1)$.
Show Solution
$(f \circ g)(2) = f(g(2)) = f(4+3) = f(7) = 14 - 1 = 13$.
$(g \circ f)(-1) = g(f(-1)) = g(-2-1) = g(-3) = 9 + 3 = 12$.
Problem 36 — Composition with Domain
Let $f(x) = \dfrac{1}{x}$ and $g(x) = x + 3$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$ with their domains.
Show Solution
$(f \circ g)(x) = \dfrac{1}{x+3}$. Domain: $x \neq -3$.
$(g \circ f)(x) = \dfrac{1}{x} + 3$. Domain: $x \neq 0$.
Problem 37 — Finding an Inverse
Find $f^{-1}(x)$ for $f(x) = \dfrac{3x - 2}{x + 1}$ and state its domain.
Show Solution
Set $y = \dfrac{3x-2}{x+1}$. Swap $x$ and $y$: $x = \dfrac{3y-2}{y+1}$. Solve for $y$:
$x(y+1) = 3y - 2 \Rightarrow xy + x = 3y - 2 \Rightarrow y(x-3) = -x - 2 \Rightarrow y = \dfrac{-x-2}{x-3} = \dfrac{x+2}{3-x}$.
$f^{-1}(x) = \dfrac{x+2}{3-x}$. Domain: $x \neq 3$ (i.e., $x \neq 3$, the VA of $f^{-1}$).
Problem 38 — Verifying Inverses
Verify algebraically that $f(x) = \dfrac{3x-2}{x+1}$ and $f^{-1}(x) = \dfrac{x+2}{3-x}$ are inverses.
Show Solution
$(f \circ f^{-1})(x) = f\!\left(\dfrac{x+2}{3-x}\right) = \dfrac{3\cdot\frac{x+2}{3-x} - 2}{\frac{x+2}{3-x}+1}$.
Numerator: $\dfrac{3(x+2) - 2(3-x)}{3-x} = \dfrac{3x+6-6+2x}{3-x} = \dfrac{5x}{3-x}$.
Denominator: $\dfrac{(x+2)+(3-x)}{3-x} = \dfrac{5}{3-x}$.
$\dfrac{5x/(3-x)}{5/(3-x)} = x$ ✓. Similarly $(f^{-1} \circ f)(x) = x$ can be verified.
Problem 39 — Inverse with Restriction
Find the inverse of $f(x) = -2(x - 3)^2 + 5$ restricted to $x \leq 3$. State domain and range of the inverse.
Show Solution
With $x \leq 3$: range of $f$ is $(-\infty, 5]$.
Swap: $x = -2(y-3)^2 + 5 \Rightarrow (y-3)^2 = \dfrac{5-x}{2} \Rightarrow y - 3 = -\sqrt{\dfrac{5-x}{2}}$ (negative root since $x \leq 3$ means $y - 3 \leq 0$).
$f^{-1}(x) = 3 - \sqrt{\dfrac{5-x}{2}}$.
Domain of $f^{-1}$: $(-\infty, 5]$. Range of $f^{-1}$: $(-\infty, 3]$.
Problem 40 — Composition (Mixed Functions, Challenge)
Let $f(x) = \sqrt{x}$ and $g(x) = 1 - x^2$. Find the domain of $(f \circ g)(x)$ and $(g \circ f)(x)$ and simplify each.
Show Solution
$(f \circ g)(x) = \sqrt{1 - x^2}$. Require $1 - x^2 \geq 0 \Rightarrow x^2 \leq 1 \Rightarrow -1 \leq x \leq 1$. Domain: $[-1,\, 1]$.
$(g \circ f)(x) = 1 - (\sqrt{x})^2 = 1 - x$. Require $x \geq 0$ (domain of $f$) and $f(x) = \sqrt{x}$ must be in domain of $g$ — which is all reals, so no additional restriction. Domain: $[0, +\infty)$.
Note: though $(g \circ f)(x) = 1 - x$ looks simple, the domain is not all reals — it is restricted by the inner function $f$.