MHF4U · Exam review

Advanced Functions Exam Review

A practice exam with questions from all 6 units, then the key facts from every lesson, unit by unit. For a deeper look at one unit, open its test review.

Unit test reviews

Practice exam

36 questions, the same number from each unit. Each one shows the lesson it comes from.

Unit 1: Polynomial Functions and Their Shapes
1
Which of the following is NOT a polynomial function?
Unit 1: Zeros, Multiplicity and Sketching
2
What are the zeros of $f (x) = (2 x - 3) (x + 5)^{2}$ ?
Unit 1: The Remainder and Factor Theorems
3
What is the remainder when $P (x) = x^{3} - 2 x^{2} + 5 x - 1$ is divided by $x - 3$ ?
Unit 1: The Remainder and Factor Theorems
4
Match each divisor to the value you substitute to find the remainder.
x − 5
x + 2
2x − 3
3x + 1
Unit 1: Families of Polynomials and Polynomial Models
5
Which polynomial belongs to the same family as $y = (x + 2) (x - 1) (x - 3)$ ?
Unit 1: Polynomial Inequalities
6
Match each inequality to its interval notation.
x > 2
x ≥ 2
x < 2
x ≤ 2
Unit 2: Reciprocal Functions
7
What is the vertical asymptote of $y = \frac{1}{3 x + 9}$ ?
Unit 2: Reciprocal Functions
8
Match each feature of f with what it produces on the graph of 1/f.
A zero of f
f(x) = 1 or f(x) = −1
A minimum of f
Very large values of f
Unit 2: Restrictions, Asymptotes and Holes
9
After simplifying $\frac{x ^{2} - 25}{x ^{2} - 3 x - 10}$ to $\frac{x + 5}{x + 2}$ , what is true at $x = 5$ ?
Unit 2: Sketching Rational Functions
10
Near the vertical asymptote of $f (x) = \frac{3}{( x - 2 ) ^{2}}$ , how does the graph behave?
Unit 2: Rational Equations and Inequalities
11
Solve $\frac{5}{x - 3} = 1$ .
Unit 2: Rational Equations and Inequalities
12
Match each situation to the correct action.
A root equals a restricted value
Every root is rejected
Solving an inequality
Endpoint is a zero of the denominator
Endpoint is a zero of the numerator in a ≥ inequality
Unit 3: Introduction to Logarithms
13
Evaluate $lo g_{3} 81$ .
Unit 3: Introduction to Logarithms
14
Rewrite $lo g_{2} 128 = 7$ in exponential form.
Unit 3: Laws of Logarithms
15
Which of these is NOT a valid logarithm law?
Unit 3: Graphs of Logarithmic Functions
16
State the domain of $y = lo g (4 - x)$ .
Unit 3: Solving Exponential and Logarithmic Equations
17
Solve $2^{x} = 50$ , correct to three decimal places.
Unit 3: Solving Exponential and Logarithmic Equations
18
Solve $lo g_{4} (x - 3) = 2$ .
Unit 4: Radian Measure
19
Convert $135°$ to radians. Give the decimal value, rounded to 2 decimals.
Unit 4: The Unit Circle and Exact Values
20
Evaluate $4 cos \frac{π}{3}$ exactly.
Unit 4: Trigonometric Graphs and Transformations
21
Find the period of $y = sin (3 x)$ . Give the decimal value, rounded to 2 decimals.
Unit 4: Trigonometric Graphs and Transformations
22
A sinusoid has a maximum of $11$ , a minimum of $3$ , and a period of $π$ . Find the value of $k$ .
Unit 4: Modelling with Sinusoidal Functions
23
Using $D (t) = 3.25 sin (\frac{2 π}{365} (t - 80)) + 12.15,$ predict the day length on day $100$ , in hours to 1 decimal place.
h
Unit 4: Reciprocal Trigonometric Functions
24
Is $csc θ$ the same thing as $sin^{- 1} θ$ ? Answer yes or no.
Unit 5: Basic Trigonometric Identities
25
Which of these statements is an IDENTITY rather than an equation?
Unit 5: Compound Angle Formulas
26
Which expansion of $cos (A + B)$ is correct?
Unit 5: Double Angle Formulas
27
Which is the correct double angle formula for sine?
Unit 5: Double Angle Formulas
28
Match each expression to its simplified form.
1. A. $cos 2 θ$
2. B. $sin 2 θ$
3. C. $tan 2 θ$
4. D. $2 cos^{2} θ$
5. E. $2 sin^{2} θ$
$2 sin θ cos θ$
$1 + cos 2 θ$
$1 - cos 2 θ$
$cos^{2} θ - sin^{2} θ$
$\frac{2 tan θ}{1 - tan ^{2} θ}$
Unit 5: Proving Identities & Solving Equations
29
Match each equation to the pair of quadrants containing its solutions on $0 \leq x < 2 π$ .
$sin x = 0.4$
$sin x = - 0.4$
$cos x = 0.4$
$cos x = - 0.4$
Unit 5: Trig Equations with Multiple Angles and Applications
30
$sin x cos \frac{π}{3} + cos x sin \frac{π}{3} = \frac{1}{2}$ simplifies to which equation?
Unit 6: Average Rate of Change
31
Which expression gives the average rate of change of $f$ from $x = a$ to $x = b$ ?
Unit 6: Instantaneous Rate of Change
32
As $h$ gets smaller and smaller, the secant line through $(a, f (a))$ and $(a + h, f (a + h))$ does what?
Unit 6: Rates of Change and the Shape of a Graph
33
For a ball with $h (t) = - 4.9 t^{2} + 19.6 t + 2$ m, the rate of change at $t = 3$ s is $- 9.8$ m/s. What does this mean?
Unit 6: Composition of Functions & Modelling
34
Let $f (x) = x^{2} + 1$ and $g (x) = 2 x - 3$ . Find $f (g (2))$ .
Unit 6: Comparing Functions and Choosing a Model
35
Order these values at $x = 20$ from smallest to largest.
1. 1 $x^{2}$
2. 2 $2^{x}$
3. 3 $x^{3}$
4. 4 $lo g_{2} x$
Unit 6: Solving Equations and Inequalities Graphically and Numerically
36
$h (x) = \frac{1}{x}$ has $h (- 1) < 0$ and $h (1) > 0$ . What can you conclude?

Key facts by unit

Unit 1: Polynomial Functions

Polynomial Functions and Their Shapes

Standard form: $f (x) = a_{n} x^{n} + \dots + a_{1} x + a_{0}$ with $a_{n} \neq = 0$ and all exponents whole numbers.
End behaviour depends only on the leading term: even degree means both ends agree, odd degree means they oppose; the sign of $a_{n}$ says which way.
A degree- $n$ polynomial has at most $n$ zeros and at most $n - 1$ turning points.
Domain is always ${x \in R}$ ; the range is ${y \in R}$ exactly when the degree is odd.
For a degree- $n$ polynomial with $x$ -values spaced by $1$ , the $n$ th finite differences are constant and equal $a \cdot n!$ .
The constant term $a_{0}$ equals $f (0)$ , the $y$ -intercept.

Zeros, Multiplicity and Sketching

A factor $(x - r)$ gives the zero $x = r$ ; a factor $(a x - b)$ gives the zero $x = \frac{b}{a}$ .
In $a (x - r_{1})^{m_{1}} \dots (x - r_{k})^{m_{k}}$ , the multiplicities add to the degree.
Odd multiplicity means the graph crosses the axis; even multiplicity means it touches and turns back.
Multiplicity $3$ or higher flattens the graph against the axis near that zero.
The $y$ -intercept is $f (0)$ — substitute $x = 0$ into the factored form directly.
Zeros alone determine the curve only up to a vertical stretch $a$ ; one extra point fixes $a$ .

The Remainder and Factor Theorems

Division statement: $P (x) = D (x) Q (x) + R (x)$ , with $de g R < de g D$ .
Remainder theorem: dividing $P (x)$ by $x - a$ leaves remainder $P (a)$ ; dividing by $a x - b$ leaves $P (\frac{b}{a})$ .
Factor theorem: $x - a$ is a factor of $P (x)$ if and only if $P (a) = 0$ .
Integer zero theorem: an integer zero of a polynomial with integer coefficients must divide the constant term.
Rational zero theorem: a rational zero $\frac{p}{q}$ has $p ∣ a_{0}$ and $q ∣ a_{n}$ .
Synthetic division: bring down, multiply by $a$ , add. Last entry is the remainder; the rest are the quotient's coefficients.

Families of Polynomials and Polynomial Models

The family with zeros $r_{1}, \dots, r_{n}$ is $y = k (x - r_{1}) \dots (x - r_{n})$ , $k \neq = 0$ ; substitute one non-zero point to find $k$ .
Irrational zeros $a \pm b$ come as a pair whose factors multiply to $(x - a)^{2} - b$ .
Even: $f (- x) = f (x)$ , symmetric in the $y$ -axis. Odd: $f (- x) = - f (x)$ , symmetric about the origin.
A polynomial with only even powers (constants included) is even; only odd powers, odd; a mixture, neither.
Open box from an $L \times W$ sheet: $V (x) = x (L - 2 x) (W - 2 x)$ with domain $0 < x < \frac{W}{2}$ (the shorter side).
Solve a target-value model by rearranging to $P (x) = 0$ , using the factor theorem, then rejecting roots outside the domain.

Polynomial Inequalities

Always rearrange to compare with $0$ before factoring. Never divide both sides by an expression containing $x$ .
The sign of a polynomial can change only at its zeros, so the zeros cut the number line into test intervals.
The sign flips across a zero of odd multiplicity and does not flip across a zero of even multiplicity.
Strict inequalities ( $<$ , $>$ ) exclude the zeros and use round brackets; non-strict ( $\leq$ , $\geq$ ) include them and use square brackets.
$\infty$ and $- \infty$ always take round brackets: $[2, \infty)$ , never $[2, \infty]$ .
Separate pieces of a solution are joined with a union symbol, as in $(- 2, 0) \cup (2, \infty)$ .

Unit 2: Rational Functions

Reciprocal Functions

The reciprocal of $f$ is $\frac{1}{f ( x )}$ , defined for every $x$ with $f (x) \neq = 0$ — always state that restriction.
Each zero of $f$ becomes a vertical asymptote of $\frac{1}{f ( x )}$ .
As $∣ f (x) ∣ \to \infty$ , $\frac{1}{f ( x )} \to 0$ : the horizontal asymptote is $y = 0$ , and the graph never touches it.
$\frac{1}{f ( x )}$ has exactly the same sign as $f (x)$ on every interval where it is defined.
Invariant points occur where $f (x) = 1$ or $f (x) = - 1$ , never where $f (x) = 0$ .
A minimum of $f$ with a non-zero value lines up with a local maximum of $\frac{1}{f ( x )}$ at the same $x$ -value.

Restrictions, Asymptotes and Holes

Restrictions are the zeros of the original denominator, found before any cancelling, and they never go away.
A denominator factor that does not cancel gives a vertical asymptote; a factor that cancels gives a hole.
The $y$ -value of a hole comes from substituting into the simplified function, never the original.
Degrees: numerator lower gives $y = 0$ ; equal gives $y = \frac{a}{b}$ from the leading coefficients; numerator one higher gives an oblique asymptote.
The oblique asymptote is the quotient of the division — discard the remainder, since it tends to $0$ .
A graph may cross a horizontal or oblique asymptote in the middle; only its end behaviour is guaranteed.

Sketching Rational Functions

$x$ -intercepts are the zeros of the simplified numerator; the $y$ -intercept is $f (0)$ when $0$ is in the domain.
Critical values for the sign chart are the $x$ -intercepts and the vertical asymptotes.
The sign of $f$ on the interval beside a vertical asymptote tells you whether that branch goes to $+ \infty$ or $- \infty$ .
Near a hole the curve is smooth and finite; only one point is missing.
To test whether the curve meets its horizontal asymptote $y = k$ , solve $f (x) = k$ .

Rational Equations and Inequalities

State the restrictions — every zero of every denominator — before solving anything.
To solve an equation, multiply every term by the LCD, then solve the resulting polynomial equation.
A root that equals a restriction is extraneous: reject it and say so. All roots rejected means "no solution".
Never multiply or cross-multiply an inequality by an expression containing the variable — its sign is unknown.
For an inequality, rewrite it as a single rational expression compared to $0$ , then build a sign chart.
Critical values come from the numerator (square brackets allowed) and the denominator (always round brackets).

Unit 3: Exponential & Logarithmic Functions

Introduction to Logarithms

Definition: $lo g_{b} x = y ⟺ b^{y} = x$ , with $b > 0$ , $b \neq = 1$ , $x > 0$ .
A logarithm is an exponent — the answer to $lo g_{b} x$ is the power you put on $b$ to make $x$ .
$lo g_{b} 1 = 0$ and $lo g_{b} b = 1$ for every valid base $b$ .
$lo g x$ with no base written means $lo g_{10} x$ , the common logarithm.
$lo g_{b} 0$ and $lo g_{b} (negative)$ are undefined, because $b^{y} > 0$ for every real $y$ .
$b^{l o g_{b} x} = x$ and $lo g_{b} b^{n} = n$ — the two operations undo each other.

Laws of Logarithms

Product law: $lo g_{b} (mn) = lo g_{b} m + lo g_{b} n$ .
Quotient law: $lo g_{b} (\frac{m}{n}) = lo g_{b} m - lo g_{b} n .$
Power law: $lo g_{b} (m^{n}) = n lo g_{b} m$ — this is the one that frees an exponent.
Change of base: $lo g_{b} x = \frac{lo g _{a} x}{lo g _{a} b}$ , usually with $a = 10$ .
There is no law for $lo g_{b} (m + n)$ or $lo g_{b} (m - n)$ — a logarithm of a sum cannot be split.
All three laws require a single common base; $lo g_{2} 5 + lo g_{3} 5$ cannot be combined.

Graphs of Logarithmic Functions

$y = lo g_{b} x$ is the inverse of $y = b^{x}$ ; its graph is the reflection of $y = b^{x}$ in the line $y = x$ .
Parent function: domain $x > 0$ , range $y \in R$ , vertical asymptote $x = 0$ , $x$ -intercept $(1, 0)$ , and it passes through $(b, 1)$ .
Transformed form: $y = a lo g_{b} (k (x - d)) + c$ — same parameter roles as every other function family.
The vertical asymptote moves to $x = d$ ; the vertical shift $c$ does not move it.
Domain of a transformed logarithm: solve $k (x - d) > 0$ . It is $x > d$ when $k > 0$ and $x < d$ when $k < 0$ .
For $b > 1$ , $y = lo g_{b} x$ has no maximum and no horizontal asymptote — it rises forever, just very slowly.

Solving Exponential and Logarithmic Equations

To free a variable from an exponent, take $lo g$ of both sides and apply the power law: $b^{x} = M \Rightarrow x = \frac{lo g M}{lo g b}$ .
If both sides can be written with a common base, match bases instead — it is faster and gives an exact answer.
$lo g_{b} A = lo g_{b} B \Rightarrow A = B$ ; and $lo g_{b} A = c \Rightarrow A = b^{c}$ .
State the domain before solving: every argument must be positive. Every candidate root must then be tested against it, and extraneous roots rejected explicitly.
Log scales: $pH = - lo g [H^{+}]$ , $β = 10 lo g (\frac{I}{I _{0}})$ with $I_{0} = 1 0^{- 12}$ W/m², and $M = lo g (\frac{I}{I _{0}})$ .
A difference of $n$ units on a Richter or pH scale means a factor of $1 0^{n}$ ; a difference of $n$ decibels means a factor of $1 0^{n /10}$ .

Unit 4: Trigonometric Functions

Radian Measure

One radian is the central angle subtending an arc equal in length to the radius: $θ = \frac{a}{r}$ .
One full revolution is $2 π$ radians, so $π rad = 180°$ and $1 rad \approx 57.3°$ .
Degrees to radians: multiply by $\frac{π}{180}$ . Radians to degrees: multiply by $\frac{180}{π}$ .
Arc length $a = r θ$ and sector area $A = \frac{1}{2} r^{2} θ$ , both with $θ$ in radians.
Angular velocity $ω = \frac{θ}{t}$ (rad/s); linear speed at radius $r$ is $v = r ω$ ; and $1$ rev/min $= 2 π$ rad/min.
Radians are unitless (a length over a length), so an angle written as a bare number is in radians.

The Unit Circle and Exact Values

On the unit circle, the point at angle $θ$ is $(cos θ, sin θ)$ , and $tan θ = \frac{sin θ}{cos θ}$ .
Because $x^{2} + y^{2} = 1$ on the circle, $sin^{2} θ + cos^{2} θ = 1$ for every angle.
Related angles: $π - θ$ in QII, $θ - π$ in QIII, $2 π - θ$ in QIV.
CAST: all positive in QI, only sine in QII, only tangent in QIII, only cosine in QIV.
Coterminal angles differ by a multiple of $2 π$ and share identical trig values.
$sin \frac{π}{6} = \frac{1}{2}$ , $cos \frac{π}{6} = \frac{3}{2}$ , $sin \frac{π}{4} = cos \frac{π}{4} = \frac{2}{2}$ , $sin \frac{π}{3} = \frac{3}{2}$ , $tan \frac{π}{4} = 1$ .

Trigonometric Graphs and Transformations

$y = sin x$ and $y = cos x$ : period $2 π$ , amplitude $1$ , axis $y = 0$ , range $[- 1, 1]$ , domain all real numbers.
$y = tan x$ : period $π$ , range all real numbers, vertical asymptotes at $x = \frac{π}{2} + nπ$ , zeros at $x = nπ$ .
In $y = a sin (k (x - d)) + c$ : amplitude $= ∣ a ∣$ , period $= \frac{2 π}{∣ k ∣}$ , phase shift $= d$ (right if positive), axis $= y = c$ .
Maximum $= c + ∣ a ∣$ and minimum $= c - ∣ a ∣$ .
From a graph: $c = \frac{max + min}{2}$ , $a = \frac{max - min}{2}$ , $k = \frac{2 π}{period}$ .
Always factor $k$ out of the bracket before reading the phase shift.

Modelling with Sinusoidal Functions

Axis $c = \frac{max + min}{2}$ ; amplitude $∣ a ∣ = \frac{max - min}{2}$ .
$k = \frac{2 π}{period}$ , with the period in the model's time units.
Start at a maximum: $+ cos$ ; at a minimum: $- cos$ ; on the axis and rising: $+ sin$ . Otherwise set $d$ to the time of a maximum and use $+ cos$ .
To predict, substitute in radian mode; to find when a value occurs, isolate the trig function, take the inverse, find the second angle in the cycle, convert to time, then add periods.
For cosine the second angle is $2 π - θ$ ; for sine it is $π - θ$ .
Interpret every parameter with units: amplitude as radius or half-range, axis as average level, period as cycle length, d as the time of a key point.

Reciprocal Trigonometric Functions

$csc θ = \frac{1}{sin θ}$ , $sec θ = \frac{1}{cos θ}$ , $cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}$ .
A reciprocal graph has a vertical asymptote wherever its base function equals zero.
Where the base function peaks at $\pm 1$ , the reciprocal touches it and turns — a local minimum above the axis, a local maximum below it.
$y = csc x$ and $y = sec x$ have period $2 π$ and range $y \leq - 1$ or $y \geq 1$ ; nothing lands strictly between $- 1$ and $1$ .
$y = cot x$ has period $π$ , range all real numbers, asymptotes at $x = nπ$ , and zeros at $x = \frac{π}{2} + nπ$ .
Reciprocals keep the sign of the base function, so CAST applies to $csc$ , $sec$ , and $cot$ unchanged.

Unit 5: Trigonometric Identities & Equations

Basic Trigonometric Identities

$csc θ = \frac{1}{sin θ}$ , $sec θ = \frac{1}{cos θ}$ , $cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}$ .
$tan θ = \frac{sin θ}{cos θ}$ — undefined wherever $cos θ = 0$ , i.e. at $θ = \frac{π}{2} + kπ$ .
$sin^{2} θ + cos^{2} θ = 1$ , and its rearrangements $sin^{2} θ = 1 - cos^{2} θ$ and $cos^{2} θ = 1 - sin^{2} θ$ .
Dividing by $cos^{2} θ$ gives $tan^{2} θ + 1 = sec^{2} θ$ ; dividing by $sin^{2} θ$ gives $1 + cot^{2} θ = csc^{2} θ$ .
$cos (- θ) = cos θ$ (even); $sin (- θ) = - sin θ$ and $tan (- θ) = - tan θ$ (odd).
$sin (\frac{π}{2} - θ) = cos θ$ and $cos (\frac{π}{2} - θ) = sin θ$ .

Compound Angle Formulas

$sin (A \pm B) = sin A cos B \pm cos A sin B$ — mixed products, sign preserved.
$cos (A \pm B) = cos A cos B \mp sin A sin B$ — like products, sign FLIPPED.
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B} .$
$sin \frac{π}{12} = \frac{6 - 2}{4}$ and $cos \frac{π}{12} = \frac{6 + 2}{4}$ .
$sin \frac{5 π}{12} = \frac{6 + 2}{4}$ and $cos \frac{5 π}{12} = \frac{6 - 2}{4}$ .
Read every formula backwards too: $sin X cos Y + cos X sin Y$ collapses to $sin (X + Y)$ .

Double Angle Formulas

$sin 2 θ = 2 sin θ cos θ$ .
$cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ — all three are equal.
$tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ .
The formulas generalize: $sin 6 x = 2 sin 3 x cos 3 x$ , because the half-angle is whatever is half of the angle in front of you.
Rearranged for proofs: $1 + cos 2 θ = 2 cos^{2} θ$ and $1 - cos 2 θ = 2 sin^{2} θ$ .

Proving Identities & Solving Equations

In a proof, work ONE side only. Moving terms across the equals sign assumes the result and is circular.
Default proof strategy: convert to sine and cosine, then look for a Pythagorean substitution, then combine over a common denominator, then factor.
In an equation you MAY rearrange freely. Get everything to one side, reduce to a single trig ratio, factor, then solve each factor.
Reference angle first, quadrants second: find the acute angle from the positive value, then place it wherever the CAST rule matches the sign.
On $0 \leq x < 2 π$ , a sine or cosine equation normally has TWO solutions, and $tan x = k$ has two solutions exactly $π$ apart.
Never divide both sides by a trig ratio — factor instead, or you will delete solutions.

Trig Equations with Multiple Angles and Applications

For $sin (k x) = c$ on $0 \leq x < 2 π$ , solve $sin θ = c$ on $0 \leq θ < 2 kπ$ , then divide every angle by $k$ .
For $θ = k x + b$ , transform the interval the same way: multiply by $k$ , then add $b$ ; undo in reverse order at the end.
A whole-number $k$ normally gives $2 k$ solutions for sine or cosine with $∣ c ∣ < 1$ ; a fraction such as $\frac{x}{2}$ shrinks the interval and can leave one solution or none.
If an equation mixes $2 x$ and $x$ , expand the double angle first: $sin 2 x = 2 sin x cos x$ ; pick $cos 2 x = 2 cos^{2} x - 1$ or $1 - 2 sin^{2} x$ to match the other ratio.
Recognize compound forms: $sin x cos a + cos x sin a = sin (x + a)$ and $cos x cos a - sin x sin a = cos (x + a) .$
In applied problems the interval comes from the context (one day, one revolution); state it first and interpret every answer with units.

Unit 6: Rates of Change & Combining Functions

Average Rate of Change

Average rate of change on $[a, b]$ is $\frac{f ( b ) - f ( a )}{b - a}$ — the slope of the secant through $(a, f (a))$ and $(b, f (b))$ .
It equals $\frac{Δ y}{Δ x}$ : change in output divided by change in input.
Units of the rate are (output units) per (input unit), e.g. m/s, dollars per item, people per year.
A positive rate means the output rose overall; negative means it fell; zero means it ended where it began.
The order of subtraction must match in numerator and denominator — both "later minus earlier" or both "earlier minus later".

Instantaneous Rate of Change

Instantaneous rate of change at $x = a$ is the slope of the tangent line to $y = f (x)$ at $(a, f (a))$ .
The difference quotient $\frac{f ( a + h ) - f ( a )}{h}$ is the slope of the secant from $a$ to $a + h$ .
Shrinking $h$ makes the secant slope approach the tangent slope; $h$ may become tiny but never exactly $0$ .
Centred difference estimate: $\frac{f ( a + h ) - f ( a - h )}{2 h}$ — usually more accurate than a one-sided estimate with the same $h$ .
Algebraically: expand, cancel the common factor of $h$ , then discard the remaining terms containing $h$ because they vanish as $h$ becomes small.
Units of an instantaneous rate are the same as for an average rate: output units per input unit.

Rates of Change and the Shape of a Graph

Positive instantaneous rate on an interval: the function is increasing there. Negative: decreasing.
At a local maximum the rate changes from positive to negative; at a local minimum, from negative to positive. The tangent is horizontal at both.
A zero rate without a sign change, as for $y = x^{3}$ at $x = 0$ , is not a turning point.
The graph is steepest where the rate is largest in size, often at a point of inflection between turning points.
For a sinusoid the rate is zero at maximums and minimums and greatest in size at the axis, where it equals $∣ a ∣∣ k ∣$ .
For an exponential function the rate is proportional to the value: rate ÷ value is the same at every point.

Combining Functions

$(f \pm g) (x) = f (x) \pm g (x)$ and $(f g) (x) = f (x) \cdot g (x)$ ; the domain of each is $D_{f} \cap D_{g}$ .
$(\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}$ , with domain $D_{f} \cap D_{g}$ minus every $x$ where $g (x) = 0$ .
Domain restrictions are read off the original functions, never off the simplified result.
A function is even if $f (- x) = f (x)$ (symmetric about the $y$ -axis) and odd if $f (- x) = - f (x)$ (symmetric about the origin).
Even $\times$ odd $=$ odd, odd $\times$ odd $=$ even, even $\times$ even $=$ even — the signs multiply.
On graphs, combine outputs vertically: add, subtract, multiply or divide the two heights at each input.

Composition of Functions & Modelling

$(f \circ g) (x) = f (g (x))$ — the inner function acts first; read the notation from the inside out.
Composition is not commutative: $f (g (x)) \neq = g (f (x))$ in general.
Domain of $f \circ g$ : all $x$ in the domain of $g$ whose output $g (x)$ also lies in the domain of $f$ .
To decompose, find the operation performed last — that becomes the outer function $f$ .
If $g$ is the inverse of $f$ , then $f (g (x)) = x$ and $g (f (x)) = x$ on the appropriate domains.
When no algebraic method exists, solve by graphing both sides or by narrowing a sign change numerically, and state the precision of the answer.

Comparing Functions and Choosing a Model

Asymptotes rule out polynomials and sinusoids; vertical asymptotes suggest rational or logarithmic, horizontal ones rational or exponential.
Exponential: domain all reals, range $y > 0$ (for $y = b^{x}$ ), horizontal asymptote. Logarithmic: domain $x > 0$ , range all reals, vertical asymptote.
Sinusoidal functions are periodic with a bounded range; polynomials have no asymptotes and end behaviour set by the leading term.
Any increasing exponential eventually exceeds any polynomial: $2^{x} > x^{3}$ for all $x > 9.94$ (approximately).
Constant nth differences signal a degree-n polynomial; a constant ratio signals an exponential.
Justify a model by naming the evidence for it and a feature that rules out the nearest alternative, then check its domain and long-run behaviour in context.

Solving Equations and Inequalities Graphically and Numerically

Solve $f (x) = g (x)$ by finding the zeros of $h (x) = f (x) - g (x)$ ; the zeros of $h$ are the $x$ -coordinates of the intersections.
$h (x) > 0$ exactly where $f (x) > g (x)$ , so the sign of $h$ between its zeros solves inequalities.
If $h$ is continuous on $[a, b]$ and $h (a)$ , $h (b)$ have opposite signs, there is at least one root in $(a, b)$ .
Bisection: test the midpoint, keep the half with the sign change; after $n$ steps the width is $\frac{b - a}{2 ^{n}}$ .
Stop when the stated accuracy is guaranteed, for example when both endpoints round to the same value.
Choose a graphing window from end behaviour so no solution is hidden off the screen, and use radians for trig equations.