Chapter 10 — Differential Calculus
Calculus is the mathematics of change. Where algebra describes static relationships, calculus describes how things move, grow, and optimise. On the NBT, calculus questions test whether you understand the meaning of a derivative — not just how to mechanically apply the power rule. A student who understands that the derivative measures the gradient of a curve at any point will outperform a student who has only memorised rules.
Topics covered: limits · first principles · power rule · sum and difference rules · simplify before differentiating · gradient of a tangent · equation of a tangent · equation of a normal · stationary points · nature of turning points · second derivative · point of inflection · cubic functions and graphs · increasing and decreasing intervals · optimisation
Section 10.1 — Limits and First Principles
The derivative is defined as a limit. The gradient of a straight line is $\dfrac{\Delta y}{\Delta x}$. For a curve, the gradient changes at every point. To find it at a specific point, we shrink the gap $h$ between two points on the curve until it approaches zero:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
This is the definition from first principles. Everything else in differentiation follows from this one idea.
Topic: Reading the first principles definition
Which expression represents the derivative of $f(x)$ from first principles?
A) $\displaystyle\lim_{h \to \infty} \frac{f(x + h) - f(x)}{h}$
B) $\displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
C) $\displaystyle\lim_{h \to 0} \frac{f(x + h) + f(x)}{h}$
D) $\displaystyle\lim_{h \to 0} \frac{f(x) - f(x + h)}{x}$
Show the worked solution
Answer: B
Explanation
Step 1 — Recall the definition.
The derivative measures the instantaneous rate of change. We find it by computing the gradient of a chord between $x$ and $x + h$, then shrinking $h$ to zero:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$
Why the distractors are wrong:
A) $h \to \infty$ gives the gradient of a chord extending infinitely far — this is not a tangent.
C) The numerator uses addition instead of subtraction — this is not the difference in function values.
D) The denominator is $x$ instead of $h$ — the gradient of the chord must be divided by the horizontal gap $h$, not by $x$.
Takeaway: The two key features of the definition are: subtraction in the numerator ($f(x+h) - f(x)$) and the limit as $h \to 0$ in the denominator $h$.
Topic: Applying first principles to $f(x) = x^2$
Use first principles to find $f'(x)$ if $f(x) = x^2$.
A) $x$
B) $2x$
C) $2x + h$
D) $x^2$
Show the worked solution
Answer: B
Explanation
Step 1 — Write the definition.
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
Step 2 — Substitute $f(x) = x^2$.
$$= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$$
Step 3 — Expand the numerator.
$$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$
Step 4 — Cancel $h$ (valid since $h \neq 0$ during the limit process).
$$= \lim_{h \to 0} (2x + h)$$
Step 5 — Apply the limit.
$$= 2x$$
Why the distractors are wrong:
A) $x$ — missing the coefficient of 2; power rule gives $2x$, not $x$.
C) $2x + h$ — this is the expression before applying the limit. Once $h \to 0$, the $h$ term vanishes.
D) $x^2$ — this is the original function, not its derivative.
Takeaway: The key step is expanding $(x+h)^2$ fully, then cancelling $h$ before taking the limit.
Topic: First principles with a linear-quadratic function
Use first principles to find $f'(x)$ if $f(x) = 3x^2 - 2x$.
A) $6x - 2$
B) $3x - 2$
C) $6x + 2$
D) $6x$
Show the worked solution
Answer: A
Explanation
Step 1 — Apply the definition.
$$f'(x) = \lim_{h \to 0} \frac{[3(x+h)^2 - 2(x+h)] - [3x^2 - 2x]}{h}$$
Step 2 — Expand.
$$= \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x}{h}$$
Step 3 — Simplify (all constant terms cancel).
$$= \lim_{h \to 0} \frac{6xh + 3h^2 - 2h}{h}$$
Step 4 — Cancel $h$.
$$= \lim_{h \to 0} (6x + 3h - 2) = 6x - 2$$
Why the distractors are wrong:
B) $3x - 2$ — halved the coefficient; $\frac{d}{dx}[3x^2] = 6x$, not $3x$.
C) $6x + 2$ — sign error on the derivative of $-2x$; $\frac{d}{dx}[-2x] = -2$.
D) $6x$ — forgot to differentiate the $-2x$ term.
Takeaway: Expand fully and collect terms in $h$ before cancelling. Every term in the original function contributes to the derivative.
Topic: Recognising the first principles form
The expression $\displaystyle\lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$ equals:
A) $x^2$
B) $2x^2$
C) $3x^2$
D) $3x^3$
Show the worked solution
Answer: C
Explanation
Step 1 — Recognise the structure.
This is the first principles definition with $f(x) = x^3$. So the expression equals $f'(x)$.
Step 2 — Differentiate $f(x) = x^3$ using the power rule (or verify via expansion).
$$f'(x) = 3x^2$$
Verification by expansion:
$$\frac{(x+h)^3 - x^3}{h} = \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = 3x^2 + 3xh + h^2 \xrightarrow{h \to 0} 3x^2$$
Why the distractors are wrong:
A) $x^2$ — missing the coefficient 3.
B) $2x^2$ — this is the derivative of $x^2$ times an extra factor; doesn't apply here.
D) $3x^3$ — the derivative reduces the power by one; $x^3$ differentiates to $x^2$, not $x^3$.
Takeaway: When you see $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, identify $f(x)$ and differentiate it directly.
Section 10.2 — Rules of Differentiation
First principles works for any function but is slow. The rules of differentiation give the same result in seconds.
| Rule | Formula |
|---|---|
| Power rule | $\dfrac{d}{dx}[x^n] = nx^{n-1}$ |
| Constant | $\dfrac{d}{dx}[c] = 0$ |
| Constant multiple | $\dfrac{d}{dx}[cf(x)] = c \cdot f'(x)$ |
| Sum/difference | $\dfrac{d}{dx}[f \pm g] = f' \pm g'$ |
Note: Always simplify the expression before differentiating. Fractions, brackets, and roots must be rewritten in $ax^n$ form first.
Topic: Power rule
If $f(x) = x^5$, then $f'(x) =$
A) $5x^4$
B) $x^4$
C) $5x^6$
D) $4x^5$
Show the worked solution
Answer: A
Explanation
Apply the power rule: multiply by the exponent, reduce the exponent by 1.
$$f'(x) = 5x^{5-1} = 5x^4$$
Why the distractors are wrong:
B) Missing the coefficient 5 — you must bring the power down as a multiplier.
C) The exponent increases — differentiation reduces the power by 1, not increases it.
D) The exponent stays the same — $5 - 1 = 4$, not 5.
Takeaway: Power rule: bring down the exponent as a coefficient, then subtract 1 from the exponent.
Topic: Differentiating a polynomial
If $g(x) = 4x^3 - 3x^2 + 7x - 5$, then $g'(x) =$
A) $12x^2 - 6x + 7$
B) $12x^2 - 6x$
C) $12x^3 - 6x^2 + 7$
D) $4x^2 - 3x + 7$
Show the worked solution
Answer: A
Explanation
Differentiate term by term:
$$\frac{d}{dx}[4x^3] = 12x^2 \qquad \frac{d}{dx}[-3x^2] = -6x \qquad \frac{d}{dx}[7x] = 7 \qquad \frac{d}{dx}[-5] = 0$$
$$g'(x) = 12x^2 - 6x + 7$$
Why the distractors are wrong:
B) Dropped the $+7$ — the derivative of $7x$ is 7, not 0.
C) Kept the original exponents — each power reduces by 1 after differentiation.
D) Didn't multiply by the exponent — forgot to apply the coefficient from the power rule.
Takeaway: Differentiate every term. Constants vanish; linear terms become constants.
Topic: Simplify before differentiating — fraction
Find $f'(x)$ if $f(x) = \dfrac{3x^3 - 6x}{3x}$.
A) $2x$
B) $x - 1$
C) $2x - 1$
D) $x^2 - 1$
Show the worked solution
Answer: A
Explanation
Step 1 — Simplify by dividing each term in the numerator by $3x$.
$$f(x) = \frac{3x^3 - 6x}{3x} = \frac{3x^3}{3x} - \frac{6x}{3x} = x^2 - 2$$
Step 2 — Differentiate.
$$f'(x) = 2x$$
Takeaway: Always simplify a fraction before differentiating. Dividing each numerator term by the denominator converts it to standard polynomial form.
Why the distractors are wrong:
B) $x - 1$ — incorrect simplification.
C) $2x - 1$ — brought a phantom constant into the derivative.
D) $x^2 - 1$ — forgot to differentiate after simplifying.
Topic: Negative exponents
Find $f'(x)$ if $f(x) = \dfrac{4}{x^3}$.
A) $\dfrac{-12}{x^4}$
B) $\dfrac{12}{x^4}$
C) $\dfrac{-4}{x^2}$
D) $\dfrac{-12}{x^2}$
Show the worked solution
Answer: A
Explanation
Step 1 — Rewrite with negative exponent.
$$f(x) = 4x^{-3}$$
Step 2 — Apply power rule.
$$f'(x) = 4 \cdot (-3) x^{-3-1} = -12x^{-4} = \frac{-12}{x^4}$$
Why the distractors are wrong:
B) Sign error — the exponent $-3$ makes the derivative negative.
C) $-4/x^2$ — reduced the power by too little; $-3 - 1 = -4$, not $-2$.
D) $-12/x^2$ — correct coefficient but wrong exponent; $x^{-4}$ means $1/x^4$, not $1/x^2$.
Takeaway: Rewrite $\frac{a}{x^n}$ as $ax^{-n}$, then apply the power rule. The derivative of $x^{-n}$ is $-nx^{-n-1}$.
Topic: Rational exponents (roots)
If $f(x) = \sqrt{x}$, then $f'(x) =$
A) $\dfrac{1}{\sqrt{x}}$
B) $\dfrac{1}{2\sqrt{x}}$
C) $2\sqrt{x}$
D) $\dfrac{-1}{2\sqrt{x}}$
Show the worked solution
Answer: B
Explanation
Step 1 — Rewrite using exponent notation.
$$f(x) = x^{\frac{1}{2}}$$
Step 2 — Apply power rule.
$$f'(x) = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2x^{\frac{1}{2}}} = \frac{1}{2\sqrt{x}}$$
Why the distractors are wrong:
A) $1/\sqrt{x}$ — missing the factor of $\frac{1}{2}$.
C) $2\sqrt{x}$ — this would be the result of integrating $\sqrt{x}$, not differentiating.
D) $-1/(2\sqrt{x})$ — sign error; the exponent $\frac{1}{2}$ is positive, so the derivative is positive.
Takeaway: Rewrite $\sqrt[n]{x^m}$ as $x^{m/n}$ before differentiating.
Topic: Expand before differentiating
Find $h'(x)$ if $h(x) = (2x - 1)^2$.
A) $2(2x-1)$
B) $8x$
C) $4x - 2$
D) $8x - 4$
Show the worked solution
Answer: D
Explanation
Step 1 — Expand the bracket first (do not use chain rule — rather expand for NBT level).
$$h(x) = (2x-1)^2 = 4x^2 - 4x + 1$$
Step 2 — Differentiate term by term.
$$h'(x) = 8x - 4$$
Why the distractors are wrong:
A) $2(2x-1) = 4x - 2$ — used the coefficient 2 instead of 4 from $(2x)^2 = 4x^2$.
B) $8x$ — forgot to differentiate the constant $+1$; derivative of $-4x$ is $-4$, giving $8x - 4$, not $8x$.
C) $4x - 2$ — differentiated $4x^2 - 4x$ incorrectly; $\frac{d}{dx}[4x^2] = 8x$, not $4x$.
Takeaway: Always expand brackets before differentiating polynomial expressions.
Section 10.3 — Tangent and Normal Lines
The derivative $f'(a)$ gives the gradient of the tangent to the curve $y = f(x)$ at $x = a$. The normal at that point is perpendicular to the tangent, so its gradient is $-\dfrac{1}{f'(a)}$.
Topic: Gradient of tangent at a point
Find the gradient of the tangent to $f(x) = x^2 - 4x + 3$ at $x = 3$.
A) $0$
B) $2$
C) $6$
D) $-2$
Show the worked solution
Answer: B
Explanation
Step 1 — Differentiate.
$$f'(x) = 2x - 4$$
Step 2 — Substitute $x = 3$.
$$f'(3) = 2(3) - 4 = 2$$
Why the distractors are wrong:
A) 0 — this is $f'(2) = 2(2) - 4 = 0$; substituted the wrong $x$-value.
C) 6 — from computing $2(3) = 6$ but forgetting to subtract the $4$ in $f'(x) = 2x - 4$.
D) $-2$ — forgot to add 4 correctly; sign error.
Takeaway: The gradient of the tangent at $x = a$ is $f'(a)$. Differentiate first, then substitute.
Topic: Finding the equation of a tangent
Find the equation of the tangent to $y = x^3$ at the point where $x = 1$.
A) $y = x + 1$
B) $y = 3x - 2$
C) $y = 3x + 1$
D) $y = x - 2$
Show the worked solution
Answer: B
Explanation
Step 1 — Find the point of tangency.
$$y(1) = 1^3 = 1 \quad \Rightarrow \quad \text{point: } (1,\ 1)$$
Step 2 — Find the gradient.
$$\frac{dy}{dx} = 3x^2 \quad \Rightarrow \quad \frac{dy}{dx}\bigg|_{x=1} = 3$$
Step 3 — Use point-gradient form.
$$y - 1 = 3(x - 1) \quad \Rightarrow \quad y = 3x - 2$$
Why the distractors are wrong:
A) Gradient of 1 — used $f(1)$ as the gradient instead of $f'(1)$.
C) $y = 3x + 1$ — used $y = mx + c$ with $c = f(1) = 1$ directly; forgot to subtract using the point.
D) Gradient of 1 — same error as A with a different intercept.
Takeaway: Three steps every time: find the point, find the gradient from $f'(a)$, write $y - y_1 = m(x - x_1)$.
Topic: Finding $x$ where the gradient equals a given value
At which $x$-value does $f(x) = x^3 - 6x^2 + 9x$ have a gradient of $9$?
A) $x = 0$ or $x = 4$
B) $x = 1$ or $x = 3$
C) $x = 0$ only
D) $x = 4$ only
Show the worked solution
Answer: A
Explanation
Step 1 — Differentiate.
$$f'(x) = 3x^2 - 12x + 9$$
Step 2 — Set equal to 9.
$$3x^2 - 12x + 9 = 9$$ $$3x^2 - 12x = 0$$ $$3x(x - 4) = 0$$
Step 3 — Solve.
$$x = 0 \quad \text{or} \quad x = 4$$
Why the distractors are wrong:
B) $x = 1$ or $x = 3$ — these are the stationary points where $f'(x) = 0$, not where $f'(x) = 9$.
C/D) Only one solution — factoring gives two solutions; both must be included.
Takeaway: Set $f'(x)$ equal to the required gradient value. Do not set it to zero unless you want stationary points.
Topic: Equation of the normal
The tangent to $y = x^2 + 1$ at $x = 2$ has gradient $4$. What is the gradient of the normal at this point?
A) $4$
B) $-4$
C) $\dfrac{1}{4}$
D) $-\dfrac{1}{4}$
Show the worked solution
Answer: D
Explanation
Step 1 — Recall the relationship between tangent and normal gradients.
If the tangent gradient is $m_t$, the normal gradient is:
$$m_n = -\frac{1}{m_t}$$
Step 2 — Substitute.
$$m_n = -\frac{1}{4}$$
Why the distractors are wrong:
A) Same as tangent — parallel lines, not perpendicular.
B) $-4$ — negative of the tangent; perpendicular requires the reciprocal AND the negative.
C) $\frac{1}{4}$ — reciprocal but forgot the negative sign.
Takeaway: Tangent gradient $m$ → normal gradient $-1/m$. Both the sign flip and the reciprocal are required.
Section 10.4 — Stationary Points and Nature
A stationary point occurs where $f'(x) = 0$ — the tangent is horizontal. There are three types:
- Local maximum: gradient changes from $+$ to $-$; $f''(x) < 0$
- Local minimum: gradient changes from $-$ to $+$; $f''(x) > 0$
- Point of inflection: gradient does not change sign; $f''(x) = 0$
Topic: Finding stationary points
Find the $x$-values of the stationary points of $f(x) = x^3 - 3x$.
A) $x = 0$ only
B) $x = 1$ and $x = -1$
C) $x = 3$ and $x = -3$
D) $x = 1$ only
Show the worked solution
Answer: B
Explanation
Step 1 — Differentiate.
$$f'(x) = 3x^2 - 3$$
Step 2 — Set $f'(x) = 0$.
$$3x^2 - 3 = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1$$
Why the distractors are wrong:
A) $x = 0$ — this is the inflection point of $f(x) = x^3$, not this function.
C) $x = \pm 3$ — forgot to divide by 3 before solving.
D) Only $x = 1$ — missed the negative root; $x^2 = 1$ has two solutions.
Takeaway: Stationary points require $f'(x) = 0$. Solve the resulting equation completely — don't miss negative roots.
Topic: Nature using the second derivative
For $f(x) = x^3 - 3x$, classify the stationary point at $x = 1$.
A) Local maximum
B) Local minimum
C) Point of inflection
D) Cannot be determined
Show the worked solution
Answer: B
Explanation
Step 1 — Find $f''(x)$.
$$f'(x) = 3x^2 - 3 \quad \Rightarrow \quad f''(x) = 6x$$
Step 2 — Substitute $x = 1$.
$$f''(1) = 6(1) = 6 > 0$$
Step 3 — Interpret. Since $f''(1) > 0$, the curve is concave up at $x = 1$ — this is a local minimum.
Why the distractors are wrong:
A) Maximum requires $f'' < 0$ (concave down).
C) Inflection requires $f'' = 0$ and a sign change — $f''(1) = 6 \neq 0$.
D) The second derivative test gives a clear answer here.
Takeaway: $f'' > 0$ → minimum (cup shape). $f'' < 0$ → maximum (cap shape). $f'' = 0$ → test is inconclusive; use sign of $f'$.
Topic: Coordinates of turning points
Find the coordinates of the local maximum of $f(x) = 2x^3 - 9x^2 + 12x - 4$.
A) $(1,\ 1)$
B) $(2,\ 0)$
C) $(1,\ 0)$
D) $(2,\ 1)$
Show the worked solution
Answer: A
Explanation
Step 1 — Differentiate and find stationary points.
$$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$$
$$x = 1 \quad \text{or} \quad x = 2$$
Step 2 — Find $f''(x)$ to classify.
$$f''(x) = 12x - 18$$
$$f''(1) = -6 < 0 \quad \Rightarrow \quad \text{local maximum at } x = 1$$ $$f''(2) = 6 > 0 \quad \Rightarrow \quad \text{local minimum at } x = 2$$
Step 3 — Find $y$-coordinate of maximum.
$$f(1) = 2 - 9 + 12 - 4 = 1$$
$$\text{Local maximum: } (1,\ 1)$$
Why the distractors are wrong:
B) $(2,\ 0)$ — this is the local minimum, not the maximum.
C) $(1,\ 0)$ — correct $x$ but wrong $y$; $f(1) = 1$, not $0$.
D) $(2,\ 1)$ — swapped the coordinates of the two turning points.
Takeaway: Find both stationary points, use $f''$ to classify, then substitute the correct $x$ to find the $y$-coordinate.
Topic: Point of inflection
At which $x$-value does $f(x) = x^3 - 6x^2 + 9x + 1$ have its point of inflection?
A) $x = 1$
B) $x = 2$
C) $x = 3$
D) $x = 0$
Show the worked solution
Answer: B
Explanation
Step 1 — Differentiate twice.
$$f'(x) = 3x^2 - 12x + 9 \quad \Rightarrow \quad f''(x) = 6x - 12$$
Step 2 — Set $f''(x) = 0$.
$$6x - 12 = 0 \quad \Rightarrow \quad x = 2$$
Step 3 — Confirm sign change (not required at NBT level but worth knowing).
$f''(1) = -6 < 0$ and $f''(3) = 6 > 0$ — sign changes, confirming inflection.
Why the distractors are wrong:
A) $x = 1$ — this is a stationary point ($f'(1) = 0$), not the inflection point.
C) $x = 3$ — also a stationary point.
D) $x = 0$ — neither a stationary point nor the inflection point for this function.
Takeaway: Point of inflection: set $f''(x) = 0$. Note: for a cubic, the inflection point is always the midpoint (in $x$) of the two turning points.
Topic: Using turning point information
A cubic function $f(x) = ax^3 + bx^2 + cx + d$ has a local maximum at $x = -1$ and a local minimum at $x = 3$. What is the $x$-coordinate of its point of inflection?
A) $x = 0$
B) $x = 1$
C) $x = 2$
D) $x = -2$
Show the worked solution
Answer: B
Explanation
Step 1 — Use the key property of cubic functions.
The point of inflection of a cubic always lies at the average (midpoint) of the $x$-coordinates of the two turning points.
Step 2 — Calculate.
$$x_{\text{inflection}} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$$
Why the distractors are wrong:
A) $x = 0$ — average of $-1$ and $1$, but 1 is not the minimum; the minimum is at $x = 3$.
C) $x = 2$ — average of 1 and 3; used the wrong pair.
D) $x = -2$ — not the midpoint of any meaningful pair.
Takeaway: For a cubic $f$, the $x$-coordinate of the inflection point = $\dfrac{x_{\text{max}} + x_{\text{min}}}{2}$.
Section 10.5 — Cubic Functions and Graphs
A cubic function has the form $f(x) = ax^3 + bx^2 + cx + d$.
- If $a > 0$: falls to the left, rises to the right
- If $a < 0$: rises to the left, falls to the right
- Has at most two turning points
- Has exactly one point of inflection
- $f'(x) > 0$: function is increasing; $f'(x) < 0$: function is decreasing
Topic: Reading a cubic graph — turning points
A cubic graph has a local maximum at $(-1,\ 4)$ and a local minimum at $(2,\ -3)$. On what interval is the function decreasing?
A) $x < -1$
B) $-1 < x < 2$
C) $x > 2$
D) $x < 2$
Show the worked solution
Answer: B
Explanation
A cubic with positive leading coefficient rises, reaches a local maximum, decreases to the local minimum, then rises again.
The function decreases between the local maximum and local minimum:
$$-1 < x < 2$$
Why the distractors are wrong:
A) $x < -1$ — the function is increasing here (before the maximum).
C) $x > 2$ — the function is increasing here (after the minimum).
D) $x < 2$ — too broad; the function is increasing for $x < -1$ and only decreasing for $-1 < x < 2$.
Takeaway: Between the maximum and minimum of a cubic, the function is decreasing. Outside this interval, it is increasing.
Topic: Linking $f$, $f'$, and $f''$
The graph of $f'(x)$ has roots at $x = 1$ and $x = 5$, and opens upward (parabola). Which statement about $f(x)$ is correct?
A) $f$ has a maximum at $x = 1$ and a minimum at $x = 5$
B) $f$ has a minimum at $x = 1$ and a maximum at $x = 5$
C) $f$ has a minimum at $x = 1$ and a minimum at $x = 5$
D) $f$ has a maximum at $x = 1$ and no turning point at $x = 5$
Show the worked solution
Answer: A
Explanation
Read the sign of $f'$. It is an upward parabola with roots at 1 and 5, so it is negative between them and positive outside.
Translate to $f$:
- At $x=1$, $f'$ goes $+ \to -$ → local maximum
- At $x=5$, $f'$ goes $- \to +$ → local minimum
Why the others are wrong:
- B is reversed — an upward parabola is negative between its roots, so $f$ decreases there.
- C claims two minima. An upward $f'$ changes sign $+\to-\to+$, giving one of each.
- D denies a turning point at $x=5$, but $f'(5)=0$ and the sign changes there.
Takeaway: $f'$ going $+ \to -$ means a maximum; $- \to +$ means a minimum. Sketch the sign of $f'$ first, then read $f$ off it.
Topic: Determining a cubic equation from its graph
A cubic graph cuts the x-axis at $x = -2$, $x = 1$, and $x = 3$, and passes through $(0,\ -6)$. What is the equation of the cubic?
A) $y = (x+2)(x-1)(x-3)$
B) $y = -(x+2)(x-1)(x-3)$
C) $y = -x(x+2)(x-1)(x-3)$
D) $y = 2(x+2)(x-1)(x-3)$
Show the worked solution
Answer: B
Explanation
Step 1 — Write the general form using the roots.
$$y = k(x+2)(x-1)(x-3)$$
Step 2 — Use the point $(0,\ -6)$ to find $k$.
$$-6 = k(0+2)(0-1)(0-3) = k(2)(-1)(-3) = 6k \quad \Rightarrow \quad k = -1$$
Step 3 — Write the equation.
$$y = -(x+2)(x-1)(x-3)$$
Why the distractors are wrong:
A) $k = 1$ gives $y = +6$ at $x = 0$, not $-6$.
C) Has an extra factor $x$ — this would be degree 4, not cubic.
D) $k = 2$ gives $y = 12$ at $x = 0$, not $-6$.
Takeaway: Use known roots to write $y = k(x - r_1)(x - r_2)(x - r_3)$, then substitute a known point to find $k$.
Topic: Intervals where $f(x) > 0$
The cubic $f(x) = (x+1)(x-2)(x-4)$ has a positive leading coefficient. For which values of $x$ is $f(x) > 0$?
A) $x < -1$ or $x > 4$
B) $-1 < x < 2$ or $x > 4$
C) $x < -1$ or $2 < x < 4$
D) $-1 < x < 2$ or $2 < x < 4$
Show the worked solution
Answer: B
Explanation
Step 1 — Identify the roots: $x = -1,\ x = 2,\ x = 4$
Step 2 — Use a sign table (positive leading coefficient means the cubic is negative, positive, negative, positive in consecutive intervals):
| Interval | Sign of $f$ |
|---|---|
| $x < -1$ | negative |
| $-1 < x < 2$ | positive |
| $2 < x < 4$ | negative |
| $x > 4$ | positive |
Step 3 — Select where $f(x) > 0$:
$$-1 < x < 2 \quad \text{or} \quad x > 4$$
Why the distractors are wrong:
A) These are where the cubic would be positive if leading coefficient were negative — reversed.
C) Selects the negative intervals.
D) Skips the interval $x > 4$.
Takeaway: Draw a sign table for each interval. Positive leading coefficient → starts negative for $x < $ smallest root.
Section 10.6 — Optimisation
Optimisation uses calculus to find the maximum or minimum value of a quantity. The method:
- Write an expression for the quantity to optimise (area, volume, profit, etc.)
- Express it in terms of one variable using any constraint given
- Differentiate and set $f'(x) = 0$
- Verify using $f''(x)$ (or context)
- Answer the question — often requires finding the actual max/min value, not just $x$
Topic: Finding the maximum of a quadratic function
The profit (in rands) from selling $x$ units is $P(x) = -2x^2 + 40x - 50$. How many units maximise the profit?
A) $x = 10$
B) $x = 20$
C) $x = 40$
D) $x = 5$
Show the worked solution
Answer: A
Explanation
Step 1 — Differentiate.
$$P'(x) = -4x + 40$$
Step 2 — Set $P'(x) = 0$.
$$-4x + 40 = 0 \quad \Rightarrow \quad x = 10$$
Step 3 — Verify it's a maximum.
$$P''(x) = -4 < 0 \quad \Rightarrow \quad \text{maximum confirmed}$$
Why the distractors are wrong:
B) $x = 20$ — twice the correct answer; common error from forgetting to divide.
C) $x = 40$ — used the coefficient of $x$ directly without differentiating.
D) $x = 5$ — half the correct value.
Takeaway: Set the derivative of the objective function to zero. Verify with $f''$ or context.
Topic: Optimisation with a constraint — fence problem
A farmer has $120\,\text{m}$ of fencing to enclose a rectangular field. One side uses a river (no fencing needed). Find the length of the side parallel to the river that maximises the area.
A) $30\,\text{m}$
B) $40\,\text{m}$
C) $60\,\text{m}$
D) $80\,\text{m}$
Show the worked solution
Answer: C
Explanation
Step 1 — Set up variables.
Let the side parallel to the river have length $y$ and the two perpendicular sides each have length $x$.
Step 2 — Write the constraint (only 3 sides need fencing).
$$2x + y = 120 \quad \Rightarrow \quad y = 120 - 2x$$
Step 3 — Write the area function.
$$A = xy = x(120 - 2x) = 120x - 2x^2$$
Step 4 — Differentiate and set to zero.
$$A'(x) = 120 - 4x = 0 \quad \Rightarrow \quad x = 30$$
Step 5 — Find $y$.
$$y = 120 - 2(30) = 60\,\text{m}$$
Why the distractors are wrong:
A) 30 m — this is the length of the perpendicular sides, not the side parallel to the river.
B) 40 m — an intermediate calculation error.
D) 80 m — $120 - 2(20)$; found $x = 20$ by error.
Takeaway: Label your variables clearly. Write the constraint, use it to eliminate one variable, then differentiate the objective function.
Topic: Box optimisation
An open box is made from a $12\,\text{cm} \times 12\,\text{cm}$ square of cardboard by cutting equal squares of side $x$ from each corner and folding up the sides. Find the value of $x$ that maximises the volume.
A) $x = 1$
B) $x = 2$
C) $x = 3$
D) $x = 4$
Show the worked solution
Answer: B
Explanation
Step 1 — Write the volume formula.
After cutting, the base has dimensions $(12 - 2x) \times (12 - 2x)$ and height $x$:
$$V = x(12 - 2x)^2$$
Step 2 — Expand.
$$V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3$$
Step 3 — Differentiate.
$$V'(x) = 144 - 96x + 12x^2 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6)$$
Step 4 — Set $V'(x) = 0$.
$$x = 2 \quad \text{or} \quad x = 6$$
Step 5 — Check validity. Since $x < 6$ (otherwise the box has no base), $x = 2$.
Why the distractors are wrong:
A) $x = 1$ — $V'(1) = 12(1-2)(1-6) = 60 \neq 0$.
C) $x = 3$ — $V'(3) = 12(1)(-3) \neq 0$.
D) $x = 6$ — this makes the base zero; physically invalid.
Takeaway: Always check that your solution is physically valid. Discard solutions outside the feasible domain.
Topic: Optimisation — rate of change interpretation
A particle moves along a straight line. Its displacement (in metres) after $t$ seconds is $s(t) = t^3 - 6t^2 + 9t$. At what time is the particle momentarily at rest?
A) $t = 1$ and $t = 3$
B) $t = 0$ and $t = 2$
C) $t = 2$ only
D) $t = 3$ only
Show the worked solution
Answer: A
Explanation
Step 1 — Recall that velocity $= \dfrac{ds}{dt}$.
The particle is at rest when velocity $= 0$.
Step 2 — Differentiate.
$$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$$
Step 3 — Set $v(t) = 0$.
$$t = 1 \quad \text{or} \quad t = 3$$
Why the distractors are wrong:
B) $t = 0$ and $t = 2$ — these come from setting $s(t) = 0$, not $s'(t) = 0$.
C/D) Only one value — the quadratic $v(t) = 0$ has two roots.
Takeaway: Velocity is the derivative of displacement. "At rest" means velocity $= 0$, so set $s'(t) = 0$.
Section 10.7 — Mixed Practice
Topic: Power rule
If $f(x) = 6x^4 - 2x^3 + x$, then $f'(x) =$
A) $24x^3 - 6x^2 + 1$
B) $24x^3 - 6x^2$
C) $24x^4 - 6x^3 + 1$
D) $6x^3 - 2x^2 + 1$
Show the worked solution
Answer: A
Explanation
$$f'(x) = 24x^3 - 6x^2 + 1$$
Derivative of $6x^4 = 24x^3$; of $-2x^3 = -6x^2$; of $x = 1$.
Topic: Stationary point x-value
At which value does $g(x) = x^2 - 8x + 3$ have its turning point?
A) $x = 3$
B) $x = 4$
C) $x = 8$
D) $x = -4$
Show the worked solution
Answer: B
Explanation
$g'(x) = 2x - 8 = 0 \Rightarrow x = 4$
Topic: Nature of turning point
If $f''(a) = -7$, the turning point at $x = a$ is a:
A) Minimum
B) Point of inflection
C) Maximum
D) Cannot be determined
Show the worked solution
Answer: C
Explanation
$f'' < 0$ means concave down — local maximum.
Topic: Gradient of tangent
The gradient of the tangent to $y = 3x^2 - x$ at $x = 2$ is:
A) $10$
B) $11$
C) $6$
D) $5$
Show the worked solution
Answer: B
Explanation
$y' = 6x - 1$; at $x = 2$: $y'(2) = 12 - 1 = 11$
Topic: Simplify before differentiating
$f(x) = \dfrac{x^4 - x^2}{x}$. Then $f'(x) =$
A) $3x^2 - 1$
B) $3x^2$
C) $4x^3 - 2x$
D) $x^3 - x$
Show the worked solution
Answer: A
Explanation
$f(x) = x^3 - x$; $f'(x) = 3x^2 - 1$
Topic: First principles
Using first principles, $f'(x)$ where $f(x) = 5x$ equals:
A) $5x$
B) $0$
C) $5$
D) $5x + 5h$
Show the worked solution
Answer: C
Explanation
$\lim_{h \to 0} \frac{5(x+h) - 5x}{h} = \lim_{h \to 0} \frac{5h}{h} = 5$
Topic: Equation of tangent
The equation of the tangent to $f(x) = x^2 - 3$ at $x = -1$ is:
A) $y = -2x - 4$
B) $y = 2x - 4$
C) $y = -2x - 2$
D) $y = -2x + 4$
Show the worked solution
Answer: A
Explanation
$f(-1) = 1 - 3 = -2$; $f'(x) = 2x$; $f'(-1) = -2$
Tangent: $y - (-2) = -2(x - (-1)) \Rightarrow y = -2x - 4$
Topic: Turning points of cubic
$f(x) = x^3 - 12x + 5$ has turning points at:
A) $x = 2$ and $x = -2$
B) $x = 4$ and $x = -4$
C) $x = 3$ and $x = -3$
D) $x = 6$ only
Show the worked solution
Answer: A
Explanation
$f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$; $x = 2$ or $x = -2$
Topic: Inflection point
The point of inflection of $f(x) = x^3 - 3x^2 + 6x - 1$ occurs at $x =$
A) $1$
B) $2$
C) $3$
D) $0$
Show the worked solution
Answer: A
Explanation
$f''(x) = 6x - 6 = 0 \Rightarrow x = 1$
Topic: Increasing/decreasing
For $f(x) = -x^3 + 3x$, the function is increasing when:
A) $-1 < x < 1$
B) $x < -1$ or $x > 1$
C) $x > 0$
D) All real $x$
Show the worked solution
Answer: A
Explanation
$f'(x) = -3x^2 + 3 = -3(x^2 - 1) = -3(x-1)(x+1)$
$f'(x) > 0$ when $(x-1)(x+1) < 0$, i.e. $-1 < x < 1$
Topic: Optimisation — maximum value
Find the maximum value of $f(x) = -x^2 + 6x + 1$.
A) $3$
B) $9$
C) $10$
D) $7$
Show the worked solution
Answer: C
Explanation
$f'(x) = -2x + 6 = 0 \Rightarrow x = 3$
$f(3) = -9 + 18 + 1 = 10$
Topic: Normal gradient
The tangent to $f(x) = x^3$ at $x = -1$ has gradient $3$. The gradient of the normal is:
A) $-3$
B) $3$
C) $\dfrac{1}{3}$
D) $-\dfrac{1}{3}$
Show the worked solution
Answer: D
Explanation
$f'(x) = 3x^2$; $f'(-1) = 3$. Normal gradient $= -\frac{1}{3}$
Topic: Reading $f'$ from a graph
A function $f$ is increasing on $(-\infty,\ 2)$ and decreasing on $(2,\ \infty)$. Which statement about $f'$ is true?
A) $f'(2) = 0$ and $f'(x) < 0$ for $x < 2$
B) $f'(2) = 0$ and $f'(x) > 0$ for $x < 2$
C) $f'(2) > 0$
D) $f'(x) < 0$ for all $x$
Show the worked solution
Answer: B
Explanation
Increasing means $f' > 0$. At the maximum, $f' = 0$.
Topic: Second derivative sign
If $f''(x) > 0$ for all $x$ in an interval, then on that interval $f$ is:
A) Decreasing
B) Concave down
C) Concave up
D) At a maximum
Show the worked solution
Answer: C
Explanation
$f'' > 0$ means $f'$ is increasing — the curve is concave up (cup shape).
Topic: Cubic with negative leading coefficient
Which of the following best describes the end behaviour of $f(x) = -2x^3 + x^2 - 5$?
A) Rises left, rises right
B) Rises left, falls right
C) Falls left, rises right
D) Falls left, falls right
Show the worked solution
Answer: B
Explanation
Negative leading coefficient ($a = -2 < 0$): for large negative $x$, $-2x^3 \to +\infty$ (rises left); for large positive $x$, $-2x^3 \to -\infty$ (falls right).
Topic: Finding $k$ given a stationary point
$f(x) = x^3 - kx$ has a stationary point at $x = 2$. Find $k$.
A) $6$
B) $4$
C) $12$
D) $2$
Show the worked solution
Answer: C
Explanation
$f'(x) = 3x^2 - k = 0$ at $x = 2$:
$3(4) - k = 0 \Rightarrow k = 12$
Topic: Optimisation — tin can
A cylindrical tin has volume $250\pi\ \text{cm}^3$. The radius that minimises total surface area satisfies:
A) $r = 5$ cm
B) $r = 10$ cm
C) $r = 2$ cm
D) $r = 25$ cm
Show the worked solution
Answer: A
Explanation
$V = \pi r^2 h = 250\pi \Rightarrow h = \frac{250}{r^2}$
$SA = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{500\pi}{r}$
$\frac{d(SA)}{dr} = 4\pi r - \frac{500\pi}{r^2} = 0$
$4r^3 = 500 \Rightarrow r^3 = 125 \Rightarrow r = 5$
Topic: Discriminant of $f'$ and number of turning points
For $f(x) = x^3 + bx + c$, how many turning points does $f$ have if $b > 0$?
A) Two turning points
B) One turning point
C) No turning points
D) Cannot be determined
Show the worked solution
Answer: C
Explanation
$f'(x) = 3x^2 + b$. Since $b > 0$, $f'(x) = 3x^2 + b \geq b > 0$ for all $x$.
$f'(x)$ never equals zero — no stationary points, therefore no turning points.
Topic: Relating graphs of $f$ and $f'$
At the point where the graph of $f'$ crosses the x-axis from below, $f$ has a:
A) Local maximum
B) Local minimum
C) Point of inflection
D) Vertical asymptote
Show the worked solution
Answer: B
Explanation
$f'$ crosses from below (negative) to above (positive) the x-axis → $f'$ changes from negative to positive → $f$ changes from decreasing to increasing → local minimum.
Topic: Chain of reasoning
If $f(x) = x^4 - 8x^2$, at how many points is the tangent to the curve horizontal?
A) 1
B) 2
C) 3
D) 4
Show the worked solution
Answer: C
Explanation
$f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2) = 0$
$x = 0,\ x = 2,\ x = -2$ — three points with horizontal tangents.
Topic: Using turning point to find coefficients
$f(x) = ax^2 + bx$ has a turning point at $(3,\ -9)$. Find $a$ and $b$.
A) $a = 1,\ b = -6$
B) $a = -1,\ b = 6$
C) $a = 1,\ b = 6$
D) $a = 2,\ b = -3$
Show the worked solution
Answer: A
Explanation
Turning point at $x = 3$: $f'(x) = 2ax + b = 0$ at $x = 3$:
$6a + b = 0 \Rightarrow b = -6a$
Turning point value $f(3) = -9$:
$9a + 3b = -9 \Rightarrow 9a + 3(-6a) = -9 \Rightarrow -9a = -9 \Rightarrow a = 1$, $b = -6$
Topic: Sign of $f''$ and concavity
For $f(x) = x^4 - 6x^2$, on which interval is $f$ concave down?
A) $-1 < x < 1$
B) $x < -1$ or $x > 1$
C) $x > 0$
D) All real $x$
Show the worked solution
Answer: A
Explanation
$f'(x) = 4x^3 - 12x$; $f''(x) = 12x^2 - 12 = 12(x^2 - 1)$
$f''(x) < 0$ when $x^2 < 1$, i.e. $-1 < x < 1$
Topic: Optimisation — area
A rectangle is inscribed in a right-angled triangle with legs $6$ and $8$. If one corner is at the right angle, find the width $x$ that maximises the rectangle's area.
A) $x = 3$
B) $x = 4$
C) $x = 2$
D) $x = 6$
Show the worked solution
Answer: B
Explanation
The hypotenuse passes through $(8, 0)$ and $(0, 6)$, so its equation is $\dfrac{x}{8} + \dfrac{y}{6} = 1$, giving $y = 6\!\left(1 - \dfrac{x}{8}\right)$.
$$A = xy = x \cdot 6\!\left(1 - \frac{x}{8}\right) = 6x - \frac{3x^2}{4}$$
$$A' = 6 - \frac{3x}{2} = 0 \quad \Rightarrow \quad x = 4$$
$y = 6(1 - 4/8) = 3$; maximum area $= 4 \times 3 = 12\,\text{cm}^2$.
Topic: Velocity and acceleration
A ball's height is $h(t) = -5t^2 + 20t + 2$ metres after $t$ seconds. What is the maximum height?
A) $20\,\text{m}$
B) $22\,\text{m}$
C) $25\,\text{m}$
D) $2\,\text{m}$
Show the worked solution
Answer: B
Explanation
$h'(t) = -10t + 20 = 0 \Rightarrow t = 2$
$h(2) = -5(4) + 20(2) + 2 = -20 + 40 + 2 = 22\,\text{m}$
Topic: Reading cubic graph features
A cubic $f$ has roots at $x = -3,\ 0,\ 2$ and a negative leading coefficient. Which is the sketch?
A) Falls left, local max between $-3$ and $0$, local min between $0$ and $2$, rises right
B) Rises left, local max between $-3$ and $0$, local min between $0$ and $2$, falls right
C) Rises left, local min between $-3$ and $0$, local max between $0$ and $2$, falls right
D) Falls left, local min between $-3$ and $0$, local max between $0$ and $2$, rises right
Show the worked solution
Answer: C
Explanation
Negative leading coefficient: rises left, falls right. The function comes from $+\infty$ on the left, decreases through $x = -3$, dips to a local minimum between $-3$ and $0$, rises through $x = 0$, peaks at a local maximum between $0$ and $2$, then decreases through $x = 2$ to $-\infty$.
Checking signs: $f(-1) = -(-1+3)(-1)(-1-2) = -(2)(-1)(-3) = -6 < 0$ and $f(1) = -(1+3)(1)(1-2) = -(4)(1)(-1) = 4 > 0$, confirming the minimum is in $(-3, 0)$ (function negative) and the maximum is in $(0, 2)$ (function positive).
Topic: Using the discriminant of $f'$
How many real stationary points does $f(x) = x^3 + 3x^2 + 5x - 1$ have?
A) 0
B) 1
C) 2
D) 3
Show the worked solution
Answer: A
Explanation
$f'(x) = 3x^2 + 6x + 5$
Discriminant of $f'$: $\Delta = 36 - 4(3)(5) = 36 - 60 = -24 < 0$
No real roots → no stationary points.
Topic: Second derivative and concavity change
The graph of $f''(x)$ crosses the x-axis at $x = 3$ (from negative to positive). This means $f(x)$ has:
A) A local maximum at $x = 3$
B) A local minimum at $x = 3$
C) A point of inflection at $x = 3$
D) A stationary point at $x = 3$
Show the worked solution
Answer: C
Explanation
$f''$ changes sign at $x = 3$ → $f$ changes concavity → point of inflection at $x = 3$.
Note: A point of inflection does not require $f'(x) = 0$.
Topic: Optimisation — combined constraint
A farmer builds a rectangular pen divided into three equal sections by two interior fences parallel to the width. Total fencing is $480\,\text{m}$. Find the width $w$ that maximises area.
A) $w = 60$
B) $w = 80$
C) $w = 120$
D) $w = 40$
Show the worked solution
Answer: A
Explanation
Two lengths ($l$) and four widths ($w$): $2l + 4w = 480 \Rightarrow l = 240 - 2w$
$A = lw = (240 - 2w)w = 240w - 2w^2$
$A' = 240 - 4w = 0 \Rightarrow w = 60$
Topic: Applying calculus to a real-world graph
The velocity of a car is $v(t) = 3t^2 - 18t + 24\ \text{m/s}$. When is the car decelerating?
A) $t < 2$
B) $t > 4$
C) $2 < t < 4$
D) $t < 3$
Show the worked solution
Answer: D
Explanation
Acceleration $a(t) = v'(t) = 6t - 18$.
Decelerating means $a(t) < 0$:
$$6t - 18 < 0 \quad \Rightarrow \quad t < 3$$
Topic: Linking all calculus concepts
$f(x) = x^3 - 3x^2 - 9x + 5$ has a local maximum at $x = -1$ and a local minimum at $x = 3$. The value of $f(-1) - f(3)$ is:
A) $10$
B) $16$
C) $32$
D) $-16$
Show the worked solution
Answer: C
Explanation
$f(-1) = -1 - 3 + 9 + 5 = 10$
$f(3) = 27 - 27 - 27 + 5 = -22$
$f(-1) - f(3) = 10 - (-22) = 32$