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Calculus

Limits, differentiation from first principles, the rules of differentiation, tangents, stationary points, concavity, sketching cubic graphs, and rate-of-change and optimisation applications. Notes, past papers and a quiz — everything for this topic is one click away.

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Grade 12 CAPS Mathematics

Calculus

Limits, differentiation from first principles, the rules of differentiation, tangents, stationary points, concavity, sketching cubic graphs, and rate-of-change and optimisation applications.

About This Topic

Calculus is the single highest-weighted Paper 1 topic in Grade 12 — worth 35±3 of 150 marks.

The big idea

Calculus is the study of change. The derivative — the core idea of this whole topic — measures how fast a quantity is changing at a single instant, not just on average. A car's speedometer is a real-life derivative: it reports your speed right now, not your average speed for the whole trip.

By the end of this lesson

You will find a derivative from first principles and by rule, find the equation of a tangent, locate and classify stationary points, use the second derivative for concavity, sketch a cubic using calculus, and apply differentiation to rate-of-change and optimisation problems.

Revision from Grade 11
Grade 11 already introduced average gradient between two points, and an intuitive sense of gradient at a single point. The next slide makes the distinction between the two completely explicit, with real numbers, before anything else in this lesson.

Average Gradient vs Instantaneous Gradient

This is the one distinction the entire topic of calculus is built on — get this clear before anything else.

Average Gradient (Grade 11)

Measures: steepness between two points on a curve

\[\dfrac{f(b)-f(a)}{b-a}\]

Picture: the secant line through both points

Instantaneous Gradient (Grade 12 — new)

Measures: steepness at one single point on a curve

\[f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\]

Picture: the tangent line at that one point

An interactive diagram: drag point B along the curveA parabola y equals x squared with a fixed point A at (1,1) and a draggable point B elsewhere on the curve, joined by a red secant line. Dragging B closer to A rotates the secant line toward the dashed tangent line at A, and the displayed average gradient approaches 2.A(1,1)Btangent
Point B: \(x=\) 3.00Average gradient A→B: 4.00Instantaneous gradient at A: 2
Drag point B along the curve, on either side of A — watch the secant line (red) rotate toward the dashed tangent line, and the average gradient reading creep toward \(2\).
The exact numbers, if you'd rather read them than drag

For \(f(x)=x^2\), compare both at \(x=1\) — average gradient, \(x=1\) to \(x=b\):

\(b=3\)\(\dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4\)
\(b=2\)\(\dfrac{4-1}{1}=3\)
\(b=1.5\)\(\dfrac{2.25-1}{0.5}=2.5\)
\(b=1.1\)\(\dfrac{1.21-1}{0.1}=2.1\)
\(b=1.01\)\(\dfrac{1.0201-1}{0.01}=2.01\)
\(b\to1\)\(\to\boxed{2}\)

As \(b\) creeps closer to \(1\), the average gradient creeps closer to \(2\) — and \(f'(1)=2(1)=2\) by the power rule, exactly matching. The instantaneous gradient is the limit the average gradient is heading toward as the second point merges into the first.

So which do I use?
If a question gives you two \(x\)-values (or two points), it wants the average gradient — use \(\dfrac{f(b)-f(a)}{b-a}\), no calculus needed. If it gives you one \(x\)-value and asks for a gradient, tangent, or rate of change at that point, it wants the instantaneous gradient — that's what this entire chapter teaches you to find.
Quick Check

A question gives you \(f(2)\) and \(f(5)\) and asks for the gradient between those two points. Which gradient does it want?

For \(f(x)=x^2\), what is the average gradient from \(x=2\) to \(x=4\)?

Limits

A limit describes the value a function gets closer and closer to, even at a point where the function itself may not be defined.

Notation
\[\lim_{x\to a}f(x)=L\]

Read as: as \(x\) gets closer to \(a\), \(f(x)\) gets closer to \(L\).

The simplest case

Most of the time, a limit is just direct substitution: \(\displaystyle\lim_{x\to2}(3x+1)=3(2)+1=7\). The interesting cases (next slide) are when direct substitution gives \(\frac00\).

Why it matters here

The derivative itself is defined as a limit — the limit of an average rate of change as the interval shrinks to a single instant. Every differentiation rule in this lesson comes from that one limit. CAPS asks only for an intuitive understanding of limits, not a formal proof-based treatment.

Want more detail? Read Siyavula — 6.1 Limits
Worked example — direct substitution Level 1

Evaluate \(\displaystyle\lim_{x\to2}(3x+1)\).

Show solution
  1. 1\(3x+1\) is defined everywhere, including at \(x=2\), so just substitute directly — no algebra needed.
  2. 2\(3(2)+1=\boxed{7}\)
Worked example — factorise the \(\frac00\) form Level 2

Evaluate \(\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}\).

Show solution
  1. 1Direct substitution gives \(\frac{0}{0}\), so factorise first: \(x^2-9=(x-3)(x+3)\).
  2. 2Rewrite: \(\dfrac{(x-3)(x+3)}{x-3}\).
  3. 3Cancel the common factor (valid since \(x\ne3\) in a limit): \(=x+3\).
  4. 4Now substitute \(x=3\): \(3+3=\boxed{6}\).
Worked example — factorise both top and bottom Level 3

Evaluate \(\displaystyle\lim_{x\to-1}\dfrac{x^2-1}{x^2+3x+2}\).

Show solution
  1. 1Direct substitution again gives \(\frac00\). This time both the numerator and denominator need factorising: \(x^2-1=(x-1)(x+1)\) and \(x^2+3x+2=(x+1)(x+2)\).
  2. 2Rewrite: \(\dfrac{(x-1)(x+1)}{(x+1)(x+2)}\)
  3. 3Cancel the common factor \((x+1)\), valid since \(x\ne-1\) in the limit: \(=\dfrac{x-1}{x+2}\)
  4. 4Substitute \(x=-1\): \(\dfrac{-1-1}{-1+2}=\dfrac{-2}{1}=\boxed{-2}\)

The Derivative From First Principles

This is the definition every other rule in this lesson is built from.

Definition
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]
Method
1

Write down \(f(x+h)\) by replacing every \(x\) with \((x+h)\).

2

Substitute into the formula and expand \(f(x+h)-f(x)\).

3

Factor out \(h\) from every remaining term, then cancel it with the \(h\) in the denominator.

4

Let \(h\to0\): any term still containing \(h\) disappears.

A secant line through two points on a curve, sliding toward the tangentA curve with a fixed point x and a second point x+h further along it, joined by a secant line; as h shrinks toward zero, the point x+h slides toward x and the secant line rotates into the horizontal tangent line shown dashed at x.xx+htangentsecant
The whole idea in one picture: the secant line through \(x\) and \(x+h\) has gradient \(\dfrac{f(x+h)-f(x)}{h}\). As \(h\to0\), the point \(x+h\) slides along the curve toward \(x\), and the secant line rotates until it becomes the tangent line at \(x\) — that rotation is exactly what the limit computes.
Why the limit?
\(\dfrac{f(x+h)-f(x)}{h}\) is the average gradient between \(x\) and \(x+h\) — the same average-gradient idea from Grade 11. As \(h\to0\), that interval shrinks to a single point, and the average gradient becomes the instantaneous gradient at \(x\). That is exactly what \(f'(x)\) means.
Want more detail? Read Siyavula — 6.2 Differentiation From First Principles
Exam scope
CAPS only examines first principles on three function types: constant functions (e.g. \(f(x)=c\)), quadratic-type polynomials (\(f(x)=ax^2+bx+c\)), and reciprocal functions of the form \(f(x)=\dfrac{a}{x}\). You will never be asked to use first principles on an unfamiliar or unlisted function type.
Worked example — constant type Level 1

Determine \(f'(x)\) from first principles if \(f(x)=7\).

Show solution
  1. 1\(f(x+h)=7\) as well — a constant function gives the same output for every input.
  2. 2\(f(x+h)-f(x)=7-7=0\).
  3. 3\(\dfrac{f(x+h)-f(x)}{h}=\dfrac{0}{h}=0\) for every \(h\ne0\).
  4. 4Let \(h\to0\): \(\boxed{f'(x)=0}\) — matching the constant rule you'll meet on the next slide.
Worked example — quadratic type Level 2

Determine \(f'(x)\) from first principles if \(f(x)=x^2-4x\).

Show solution
  1. 1\(f(x+h)=(x+h)^2-4(x+h)=x^2+2xh+h^2-4x-4h\).
  2. 2\(f(x+h)-f(x)=\left(x^2+2xh+h^2-4x-4h\right)-\left(x^2-4x\right)=2xh+h^2-4h\).
  3. 3Factor out \(h\): \(=h(2x+h-4)\). Divide by \(h\): \(=2x+h-4\).
  4. 4Let \(h\to0\): \(\boxed{f'(x)=2x-4}\).
Worked example — reciprocal type Level 3

Determine \(f'(x)\) from first principles if \(f(x)=-\dfrac{1}{x}\).

Show solution
  1. 1\(f(x+h)-f(x)=-\dfrac{1}{x+h}-\left(-\dfrac{1}{x}\right)=\dfrac{1}{x}-\dfrac{1}{x+h}\).
  2. 2Use a common denominator, \(x(x+h)\): \(\dfrac{1}{x}=\dfrac{x+h}{x(x+h)}\) and \(\dfrac{1}{x+h}=\dfrac{x}{x(x+h)}\), so \(\dfrac{1}{x}-\dfrac{1}{x+h}=\dfrac{(x+h)-x}{x(x+h)}=\dfrac{h}{x(x+h)}\).
  3. 3Divide by \(h\): \(\dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{x(x+h)}\).
  4. 4Let \(h\to0\): \(\boxed{f'(x)=\dfrac{1}{x^2}}\).

Source: Free State Grade 12 Preparatory Examination, September 2025, Paper 1.

Quick Check

Evaluate \(\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}\).

Which of these is NOT a function type CAPS lists for first-principles differentiation (constants, \(ax^2+bx+c\)-style polynomials, or \(a/x\))?

Rules of Differentiation

In practice, you will use these rules directly rather than first principles — but they all come from the same limit definition. \(\dfrac{d}{dx}[\,\cdot\,]\) is just another way of writing "the derivative of" — the same idea as \(f'(x)\), used when the function is written directly in terms of \(x\) rather than named \(f\).

Power rule
\[\frac{d}{dx}\left[x^n\right]=nx^{n-1}\]

e.g. \(\dfrac{d}{dx}\left[x^5\right]=5x^4\).

Constant and scalar rules
\[\frac{d}{dx}\left[k\right]=0 \qquad \frac{d}{dx}\left[k\cdot f(x)\right]=k\cdot f'(x)\]
Sum and difference
\[\frac{d}{dx}\left[f(x)\pm g(x)\right]=f'(x)\pm g'(x)\]

Differentiate term by term.

Before you differentiate

Rewrite roots and fractions as powers of \(x\) first: \(\sqrt{x}=x^{\frac{1}{2}}\) and \(\dfrac{1}{x^n}=x^{-n}\). The power rule only applies once every term is in the form \(x^n\).

Why trust the shortcut?
Earlier, first principles gave \(f'(x)=2x-4\) for \(f(x)=x^2-4x\) the long way. Using the power rule instead: \(\dfrac{d}{dx}[x^2]=2x\) and \(\dfrac{d}{dx}[-4x]=-4\), so \(f'(x)=2x-4\) — the same answer, instantly. That agreement is exactly why the rules are safe to use instead of first principles every time.
Want more detail? Read Siyavula — 6.3 Rules For Differentiation
No product or quotient rule
CAPS Grade 12 does not include a product rule or a quotient rule. If two expressions are multiplied together, expand the brackets into a sum of separate terms first, then differentiate term by term with the rules above. Getting this wrong: for \(f(x)=(x+1)(x+2)=x^2+3x+2\), the correct derivative is \(f'(x)=2x+3\). Differentiating each bracket separately and multiplying — \((1)(1)=1\) — gives a completely different, wrong answer.
Worked example — positive whole-number powers Level 1

Determine \(f'(x)\) if \(f(x)=x^4+3x^2-5x+7\).

Show solution
  1. 1Differentiate term by term using the power rule (\(\frac{d}{dx}[x^n]=nx^{n-1}\)) and the fact that a constant's derivative is \(0\).
  2. 2\(\dfrac{d}{dx}[x^4]=4x^3\), \(\quad\dfrac{d}{dx}[3x^2]=6x\), \(\quad\dfrac{d}{dx}[-5x]=-5\), \(\quad\dfrac{d}{dx}[7]=0\).
  3. 3\(\boxed{f'(x)=4x^3+6x-5}\)
Worked example — negative and fractional powers Level 3

Determine \(f'(x)\) if \(f(x)=2x^3-\dfrac{5}{x^2}+4\sqrt{x}\).

Show solution
  1. 1Rewrite every term as a power of \(x\): \(f(x)=2x^3-5x^{-2}+4x^{\frac{1}{2}}\).
  2. 2Differentiate term by term using the power rule.
  3. 3\(\dfrac{d}{dx}\left[2x^3\right]=6x^2\), \(\quad\dfrac{d}{dx}\left[-5x^{-2}\right]=10x^{-3}\), \(\quad\dfrac{d}{dx}\left[4x^{\frac12}\right]=2x^{-\frac12}\).
  4. 4So \(f'(x)=6x^2+10x^{-3}+2x^{-\frac12}\), which can be written \(\boxed{f'(x)=6x^2+\dfrac{10}{x^3}+\dfrac{2}{\sqrt{x}}}\).
Quick Check

Rewrite \(\dfrac{1}{x^3}\) as a power of \(x\), ready to differentiate.

If \(f(x)=5x^4-2\sqrt{x}\), what is \(f'(x)\)?

Equation of a Tangent

The derivative at a point is the gradient of the tangent line at that exact point.

A tangent line touching a curve at one pointA smooth curve labelled y equals f of x, with a straight tangent line touching the curve at point P.y = f(x)Ptangent
Key idea: the tangent at P has gradient \(f'(x_P)\) — find the point, find the gradient with the derivative, then use the point-gradient form of a line.
Method
1

Find the y-coordinate of the point by substituting \(x\) into \(f(x)\).

2

Find the gradient by substituting the same \(x\)-value into \(f'(x)\).

3

Substitute the point and gradient into \(y-y_1=m(x-x_1)\).

Want more detail? Read Siyavula — 6.4 Equation Of A Tangent To A Curve
Worked example — simplest case Level 1

Determine the equation of the tangent to \(f(x)=x^2\) at the point where \(x=3\).

Show solution
  1. 1Find the point: \(f(3)=9\), so the point is \((3,9)\).
  2. 2Find the gradient: \(f'(x)=2x\), so \(f'(3)=6\).
  3. 3Substitute into \(y-y_1=m(x-x_1)\): \(y-9=6(x-3)\).
  4. 4Simplify: \(y=6x-18+9=\boxed{6x-9}\).
Worked example Level 2-3

Determine the equation of the tangent to \(f(x)=x^2-4x\) at the point where \(x=5\).

Show solution
  1. 1Find the point: \(f(5)=25-20=5\), so the point is \((5,5)\).
  2. 2Find the gradient using \(f'(x)=2x-4\) (found earlier): \(f'(5)=2(5)-4=6\).
  3. 3Substitute into \(y-y_1=m(x-x_1)\): \(y-5=6(x-5)\).
  4. 4Simplify: \(y=6x-30+5=\boxed{6x-25}\).
Worked example — gradient already given Level 2

Determine the equation of the tangent to \(f(x)=x^2-3x\) that is parallel to the line \(y=x-5\).

Show solution
  1. 1The line is already in \(y=mx+c\) form, so its gradient is \(m=1\).
  2. 2A tangent parallel to this line must have the same gradient, so set \(f'(x)=1\): \(2x-3=1 \Rightarrow x=2\).
  3. 3Find the point: \(f(2)=4-6=-2\), so the point is \((2,-2)\).
  4. 4Substitute into \(y-y_1=m(x-x_1)\): \(y-(-2)=1(x-2) \Rightarrow \boxed{y=x-4}\).
Worked example — tangent parallel to a line Level 3

Determine the equation of the tangent to \(f(x)=x^2-2x-3\) that is parallel to the line \(4x-2y+6=0\).

Show solution
  1. 1Rearrange the line into \(y=mx+c\) form first: \(4x-2y+6=0 \Rightarrow -2y=-4x-6 \Rightarrow y=2x+3\), so the line's gradient is \(m=2\) — not the \(4\) sitting in front of \(x\) in the original equation.
  2. 2A tangent parallel to this line must have the same gradient, so set \(f'(x)=2\).
  3. 3\(f'(x)=2x-2=2 \Rightarrow x=2\).
  4. 4Find the point: \(f(2)=4-4-3=-3\), so the point is \((2,-3)\).
  5. 5Substitute into \(y-y_1=m(x-x_1)\): \(y-(-3)=2(x-2) \Rightarrow \boxed{y=2x-7}\).
Common trap
A line given in general form must be rearranged into \(y=mx+c\) before you can read off its gradient. Here, a student who grabs the \(4\) in front of \(x\) without rearranging would set \(f'(x)=4\) instead of \(2\), solve \(2x-2=4\) to get \(x=3\), and land on the completely different (wrong) tangent \(y=4x-12\) — a real, different answer, not just a technicality.
Quick Check

The gradient of the tangent to a curve at point P equals:

Determine the equation of the tangent to \(f(x)=x^2-4x\) at \(x=0\).

For \(f(x)=x^2-2x-3\), at which \(x\)-value is the tangent to \(f\) parallel to the line \(y=4x+5\)?

Stationary Points

A stationary point is where the gradient of the curve is exactly zero — the tangent there is horizontal.

Finding them
\[f'(x)=0\]

Solve this equation for \(x\), then substitute back into \(f(x)\) to get the \(y\)-coordinate.

Classifying them
  • Local minimum: \(f'(x)\) changes from negative to positive, or \(f''(x)>0\).
  • Local maximum: \(f'(x)\) changes from positive to negative, or \(f''(x)<0\).
What is f″(x)?
\(f''(x)\), read "f double-prime of x", is the second derivative — differentiate \(f'(x)\) one more time. The next section explains this properly; for now, just know \(f''(x)>0\) means the curve bends upward like a cup (a minimum sits there), and \(f''(x)<0\) means it bends downward like a cap (a maximum sits there).
Want more detail? Read Siyavula — 6.5 Second Derivative
Worked example — derivatives already given Level 1

Given \(f'(x)=2x-6\) and \(f''(x)=2\), determine the \(x\)-value of the stationary point and classify it.

Show solution
  1. 1Set \(f'(x)=0\): \(2x-6=0\Rightarrow\boxed{x=3}\)
  2. 2\(f''(x)=2\) is always positive, so \(f''(3)=2>0\)
  3. 3\(\boxed{x=3\text{ is a local minimum}}\)
Worked example — find f′(x) and f″(x) yourself Level 3

Determine the stationary point of \(f(x)=x^2-4x\) and classify it.

Show solution
  1. 1Set \(f'(x)=0\): \(2x-4=0 \Rightarrow x=2\).
  2. 2Find the \(y\)-coordinate: \(f(2)=4-8=-4\). The stationary point is \((2,-4)\).
  3. 3Classify using the second derivative: \(f''(x)=2\), which is always positive.
  4. 4Since \(f''(2)=2>0\), \(\boxed{(2,-4)\text{ is a local minimum}}\).
The parabola y equals x squared minus 4x showing its turning pointA smooth upward parabola crossing the x-axis at x=0 and x=4, with its lowest point (turning point) marked at (2,-4).04(2,−4)
Reading the graph: the x-intercepts \((0,0)\) and \((4,0)\) come from \(f(x)=0\); the turning point \((2,-4)\) is exactly the stationary point found by calculus above — the two methods agree.
Worked example — classify using a sign diagram Level 3

Determine and classify the stationary points of \(f(x)=x^3-3x\) using the sign of \(f'(x)\).

Show solution
  1. 1\(f'(x)=3x^2-3=3(x-1)(x+1)=0 \Rightarrow x=-1\) or \(x=1\).
  2. 2Test the sign of \(f'(x)=3(x-1)(x+1)\) in each region using a value from that region: \(f'(-2)=3(-3)(-1)=9>0\), \(\quad f'(0)=3(-1)(1)=-3<0\), \(\quad f'(2)=3(1)(3)=9>0\).
  3. 3So \(f'\) is \(+\) for \(x<-1\), \(-\) for \(-1<x<1\), then \(+\) again for \(x>1\).
  4. 4At \(x=-1\): \(f'\) changes \(+\to-\), so \(f(-1)=-1+3=2\) is a \(\boxed{\text{local maximum }(-1,2)}\).
  5. 5At \(x=1\): \(f'\) changes \(-\to+\), so \(f(1)=1-3=-2\) is a \(\boxed{\text{local minimum }(1,-2)}\).
Either test works
This sign-diagram (first-derivative) test and the second-derivative test always agree — use whichever the question asks for, or whichever is faster. Here \(f''(x)=6x\): \(f''(-1)=-6<0\) (max) and \(f''(1)=6>0\) (min), confirming the same answer.

The Second Derivative and Concavity

The second derivative measures how the gradient itself is changing — it tells you which way the curve bends.

A curve changing concavity at a point of inflectionAn S-shaped curve labelled y equals g of x, marked with a point of inflection where the curve changes from bending one way to bending the other way.y = g(x)point of inflection
Key idea: the point of inflection is where \(f''(x)=0\) and the curve switches from bending one way to bending the other.
Second derivative
\[f''(x)=\frac{d}{dx}\left[f'(x)\right]\]
Reading concavity
  • \(f''(x)>0\): concave up (curve bends upward, like a cup)
  • \(f''(x)<0\): concave down (curve bends downward, like a cap)
  • \(f''(x)=0\): possible point of inflection
Want more detail? Read Siyavula — 6.5 Second Derivative
Worked example — second derivative already given Level 1

Given \(h''(x)=6x-12\), determine the \(x\)-value of the point of inflection and state whether \(h\) is concave up or down for \(x>2\).

Show solution
  1. 1Set \(h''(x)=0\): \(6x-12=0\Rightarrow\boxed{x=2}\)
  2. 2For \(x>2\), e.g. \(x=3\): \(h''(3)=18-12=6>0\)
  3. 3\(h''(x)>0\) means \(\boxed{\text{concave up}}\) for \(x>2\)
Worked example — find g″(x) yourself first Level 3-4

Given \(g(x)=x^3-3x^2-9x+5\), determine the coordinates of the point of inflection and describe the concavity on either side of it.

Show solution
  1. 1\(g'(x)=3x^2-6x-9\), so \(g''(x)=6x-6\).
  2. 2Set \(g''(x)=0\): \(6x-6=0 \Rightarrow x=1\).
  3. 3Find \(y\): \(g(1)=1-3-9+5=-6\). Point of inflection: \(\boxed{(1,-6)}\).
  4. 4For \(x<1\), e.g. \(x=0\): \(g''(0)=-6<0\), so the curve is concave down. For \(x>1\), e.g. \(x=2\): \(g''(2)=6>0\), so the curve is concave up.
The cubic graph of g(x)=x cubed minus 3x squared minus 9x plus 5A cubic curve with a local maximum at (-1,10), a local minimum at (3,-22), and a point of inflection at (1,-6) where the concavity switches from down to up.(−1,10)(3,−22)(1,−6)
Reading the graph: the curve is concave down (bending like a cap) up to \(x=1\), then concave up (bending like a cup) after — exactly matching the point of inflection found above.
Quick Check

For \(f(x)=x^2-4x\), classify the stationary point at \(x=2\) using \(f''(x)\).

If \(f''(x)\) changes sign at \(x=a\) (and \(f''(a)=0\)), then \(x=a\) is a:

Sketching a Cubic With Calculus

Four features are enough to sketch almost any cubic accurately.

FeatureHow to find it
y-interceptSubstitute \(x=0\)
x-intercept(s)Factorise \(f(x)=0\) (factor theorem, then a quadratic factor)
Stationary pointsSolve \(f'(x)=0\), classify with \(f''(x)\)
Point of inflectionSolve \(f''(x)=0\)
Note
Plot the intercepts and stationary points first, then draw one smooth curve through them — a cubic changes direction at most twice. CAPS also expects you to describe the end behaviour: as \(x\to+\infty\), \(y\to+\infty\) for a positive leading coefficient (and the mirror image for a negative one).
Want more detail? Read Siyavula — 6.6 Sketching Graphs
Worked example — only two terms Level 1

Given \(f(x)=x^3-3x^2\), determine the intercepts and stationary points, then describe the shape of the graph.

Show solution
  1. 1y-intercept: \(f(0)=0\).
  2. 2x-intercepts: every term shares a factor of \(x^2\): \(f(x)=x^2(x-3)\), so \(x=0\) (a double root) or \(x=3\).
  3. 3Stationary points: \(f'(x)=3x^2-6x=3x(x-2)=0\Rightarrow x=0\) or \(x=2\)
  4. 4\(f(0)=0\) and \(f(2)=8-12=-4\)
  5. 5\(f''(x)=6x-6\): \(f''(0)=-6<0\) (local maximum at \((0,0)\)); \(f''(2)=6>0\) (local minimum at \((2,-4)\))
  6. 6The graph touches the x-axis at the local maximum \(\boxed{(0,0)}\) — matching the double root there — falls to the local minimum \((2,-4)\), then rises to cross the axis again at \(x=3\).
Worked example — common factor only, no factor theorem needed Level 2

Given \(f(x)=x^3-6x^2+9x\), determine the intercepts and stationary points, then describe the shape of the graph.

Show solution
  1. 1y-intercept: \(f(0)=0\).
  2. 2x-intercepts: every term shares a factor of \(x\): \(f(x)=x(x^2-6x+9)=x(x-3)^2\), so \(x=0\) or \(x=3\) (a double root) — no factor theorem trial-and-error needed here.
  3. 3Stationary points: \(f'(x)=3x^2-12x+9=3(x-1)(x-3)=0\Rightarrow x=1\) or \(x=3\)
  4. 4\(f(1)=1-6+9=4\) and \(f(3)=27-54+27=0\)
  5. 5\(f''(x)=6x-12\): \(f''(1)=-6<0\) (local maximum at \((1,4)\)); \(f''(3)=6>0\) (local minimum at \((3,0)\))
  6. 6The graph rises from the y-intercept \((0,0)\) to a local maximum at \((1,4)\), then falls to touch the x-axis at the local minimum \(\boxed{(3,0)}\) — matching the double root there — then rises again.
Worked example — factor theorem needed Level 4

Given \(f(x)=x^3-3x^2-9x+27\), determine the intercepts and stationary points, then describe the shape of the graph.

Show solution
  1. 1y-intercept: \(f(0)=27\).
  2. 2x-intercepts: \(f(x)=x^3-3x^2-9x+27\) factorises as \((x-3)^2(x+3)\), so \(x=3\) (a double root) or \(x=-3\).
  3. 3Stationary points: \(f'(x)=3x^2-6x-9=3(x-3)(x+1)=0 \Rightarrow x=3\) or \(x=-1\).
  4. 4\(f(-1)=-1-3+9+27=32\) and \(f(3)=27-27-27+27=0\).
  5. 5Since \(f''(x)=6x-6\): \(f''(-1)=-12<0\) (local maximum at \((-1,32)\)); \(f''(3)=12>0\) (local minimum at \((3,0)\)).
  6. 6So the graph rises to a local maximum at \((-1,32)\), falls to touch the x-axis at the local minimum \(\boxed{(3,0)}\) — matching the double root there — then rises again.
The cubic graph of f(x)=x cubed minus 3x squared minus 9x plus 27A cubic curve crossing the x-axis at x=-3, rising to a local maximum at (-1,32), falling to touch the x-axis at a double root local minimum (3,0), then rising again.−327(−1,32)(3,0)
Reading the graph: local maximum \((-1,32)\), local minimum \((3,0)\) — the curve just touches the x-axis there since \(x=3\) is a double root, exactly matching the calculus above.
Worked example — three distinct roots Level 4

Given \(f(x)=x^3-3x^2-6x+8\), determine the intercepts and stationary points, then sketch the graph.

Show solution
  1. 1y-intercept: \(f(0)=8\).
  2. 2x-intercepts: test \(x=1\): \(f(1)=1-3-6+8=0\), so \((x-1)\) is a factor. Dividing gives \(f(x)=(x-1)(x^2-2x-8)=(x-1)(x-4)(x+2)\), so \(x=1\), \(x=4\), or \(x=-2\) — three distinct x-intercepts.
  3. 3Stationary points: \(f'(x)=3x^2-6x-6=3(x^2-2x-2)\). This quadratic doesn't factorise neatly, so use the quadratic formula: \(x=\dfrac{2\pm\sqrt{4+8}}{2}=1\pm\sqrt3\).
  4. 4Substituting back (exact values): local maximum \(\boxed{(1-\sqrt3,\ 6\sqrt3)}\approx(-0.73,10.39)\), local minimum \(\boxed{(1+\sqrt3,\ -6\sqrt3)}\approx(2.73,-10.39)\).
  5. 5Point of inflection: \(f''(x)=6x-6=0 \Rightarrow x=1\), and \(f(1)=0\) — the inflection point \((1,0)\) happens to coincide with the middle x-intercept here.
The cubic graph of f(x)=x cubed minus 3x squared minus 6x plus 8, with three distinct x-interceptsA cubic curve crossing the x-axis at three separate points, x=-2, x=1 and x=4, with a local maximum before the middle intercept and a local minimum after it.−214(1−√3, 6√3)(1+√3, −6√3)
Reading the graph: three separate crossings of the x-axis (not a repeated root) — the curve rises, turns at the local maximum, falls through all three intercepts' region, turns at the local minimum, then rises again.
Worked example — three distinct roots, no repeated factor Level 2

A cubic function \(f\) has x-intercepts at \(x=-1\), \(x=1\) and \(x=2\), and passes through \((0,4)\). Determine \(f(x)\).

Show solution
  1. 1Three distinct roots mean \(f(x)=a(x+1)(x-1)(x-2)\) for some constant \(a\).
  2. 2Use the given point \((0,4)\) to find \(a\): \(f(0)=a(1)(-1)(-2)=2a=4 \Rightarrow a=2\).
  3. 3So \(f(x)=2(x+1)(x-1)(x-2)\).
  4. 4Expand: \(2(x^2-1)(x-2)=2(x^3-2x^2-x+2)\).
  5. 5\(\boxed{f(x)=2x^3-4x^2-2x+4}\).
Worked example — reverse problem: find the equation Level 4

A cubic function \(f\) has x-intercepts at \(x=-1\) and \(x=3\), where \(x=3\) is a repeated root (the graph touches the x-axis there), and passes through \((0,-9)\). Determine \(f(x)\).

Show solution
  1. 1A single root at \(x=-1\) and a repeated (double) root at \(x=3\) means \(f(x)=a(x+1)(x-3)^2\) for some constant \(a\).
  2. 2Use the given point \((0,-9)\) to find \(a\): \(f(0)=a(1)(9)=9a=-9 \Rightarrow a=-1\).
  3. 3So \(f(x)=-(x+1)(x-3)^2\).
  4. 4Expand: \(-(x+1)(x^2-6x+9)=-(x^3-6x^2+9x+x^2-6x+9)=-(x^3-5x^2+3x+9)\).
  5. 5\(\boxed{f(x)=-x^3+5x^2-3x-9}\).
Working backwards
Whenever you're given a repeated root, write the factor squared from the start — a repeated root always means the graph touches the axis there rather than crossing it, which is exactly what a squared factor produces.
Quick Check

A cubic has a local maximum at \(x=-1\) and a local minimum at \(x=3\). Between \(x=-1\) and \(x=3\), is the function increasing or decreasing?

For \(f(x)=x^3-3x^2-9x+27\) (the worked example above), what is the y-intercept?

Reading the Graph of f′

A very common exam question shows you the graph of \(f'(x)\) — never \(f(x)\) itself — and asks you to describe \(f\). No equation is needed if you know what each feature means.

Turning points of f

Wherever the \(f'(x)\) graph crosses or touches the x-axis, \(f'(x)=0\) there — so that x-value is a stationary point of \(f\).

Increasing / decreasing

Where the \(f'(x)\) graph is above the x-axis, \(f'(x)>0\) so \(f\) is increasing. Where it's below, \(f'(x)<0\) so \(f\) is decreasing.

Nature of each turning point

If \(f'\) changes \(+\to-\) through a root, \(f\) has a local maximum there. If \(f'\) changes \(-\to+\), a local minimum. If \(f'\) only touches the axis (a repeated root), \(f\) has a stationary point of inflection — not a max or min.

Concavity of f (the tricky part)

This needs \(f''\), but you only have the graph of \(f'\) — so read it from how the \(f'\) curve itself behaves: where the \(f'\) graph is rising, \(f\) is concave up; where the \(f'\) graph is falling, \(f\) is concave down. The x-value where the \(f'\) graph turns (its own vertex) is \(f\)'s point of inflection.

An annotated graph of f prime showing how to read all four rules at onceThe graph of f'(x)=x^2-4: an upward parabola crossing the x-axis at x=-2 and x=2, with its own vertex at (0,-4). The region is shaded green where the curve is above the x-axis (f increasing) and amber where it is below (f decreasing). The left root is labelled local max of f, the right root local min of f, and the vertex is labelled f's point of inflection.LOCAL MAXLOCAL MIN−22(0,−4)f's point of inflectiony = f ′(x)
All four rules on one picture: green bands (\(f'>0\)) are where \(f\) is increasing, amber (\(f'<0\)) is where \(f\) is decreasing. The left root is a local max of \(f\) because \(f'\) changes \(+\to-\) there; the right root is a local min because \(f'\) changes \(-\to+\). The curve's own vertex — not a root — is \(f\)'s point of inflection, since that is where \(f'\) itself turns from falling to rising.
Common mistake
Do not confuse a root of \(f'(x)\) (a turning point of \(f\)) with a root of \(f(x)\) itself (an x-intercept of \(f\)) — without an equation for \(f\), there is no way to find \(f\)'s own x-intercepts at all. And for concavity, look at the turning point of the f′ curve, not its roots — those are two completely different features of the same picture.

The Whole Picture: f, f′ and f″ Together

One example, \(f(x)=x^3-3x\), plotted with its derivative and second derivative on the same axes — watch the same x-value matter to all three at once.

The graphs of f, f prime, and f double-prime for f(x)=x cubed minus 3x, plotted togetherThree curves sharing the same horizontal axis and a shared zero line: the cubic f(x)=x^3-3x in blue, its derivative f'(x)=3x^2-3 in green (shown at reduced scale as x^2-1 so it fits alongside f), and its second derivative f''(x)=6x in red (shown at reduced scale as x, a straight line through the origin). Vertical dashed guide lines at x=-1, x=0 and x=1 show that f's local maximum at x=-1 lines up with a root of f', f's local minimum at x=1 lines up with the other root of f', and f's point of inflection at x=0 lines up with both f' reaching its own lowest point and f'' crossing zero.x = −1x = 0x = 1f(x)f'(x)f''(x)
Reading the alignment: \(f'\) and \(f''\) are shown at a reduced vertical scale so all three fit together — only their zero-crossings and sign matter here, not their exact height. At \(x=-1\): \(f\) has a local maximum and \(f'\) crosses zero. At \(x=1\): \(f\) has a local minimum and \(f'\) crosses zero again. At \(x=0\): \(f\) has its point of inflection, which is exactly where \(f'\) reaches its own lowest point and where \(f''\) crosses zero — three different statements about the same x-value.
Why this matters
This is the whole chain in one picture: a feature of \(f\) (turning point, inflection) always shows up as a feature of \(f'\) or \(f''\) at the same x-value — a root of \(f'\) is a turning point of \(f\); a root of \(f''\) is a point of inflection of \(f\); and the turning point of \(f'\) itself is also \(f\)'s point of inflection. Three different ways of describing the same location.
Worked example — the simplest possible graph Level 1

The graph of \(f'(x)\) lies entirely above the \(x\)-axis for every value of \(x\). What can you conclude about \(f\)?

Show solution
  1. 1\(f'(x)\) is never zero and never negative — it's always positive.
  2. 2\(f'(x)>0\) everywhere means \(\boxed{f\text{ is increasing for every value of }x}\), with no stationary points at all.
Worked example — one fact at a time Level 2

The graph of \(f'(x)\) crosses the \(x\)-axis at \(x=3\), with \(f'(x)>0\) for \(x<3\) and \(f'(x)<0\) for \(x>3\). What kind of stationary point does \(f\) have at \(x=3\)?

Show solution
  1. 1\(f'(3)=0\) confirms \(x=3\) is a stationary point of \(f\).
  2. 2\(f'\) changes from positive to negative as \(x\) passes through \(3\), so \(f\) changes from increasing to decreasing there.
  3. 3\(\boxed{f\text{ has a local maximum at }x=3}\)
Worked example — graph interpretation, no equation for f Level 4

The graph below shows \(f'(x)=x^2-4\), the derivative of a function \(f\). Use only the graph to determine: (a) the stationary points of \(f\); (b) where \(f\) is increasing; (c) the point of inflection of \(f\); (d) where \(f\) is concave up.

Show solution
  1. 1(a) \(f'(x)=0\) where the graph crosses the x-axis: \(x=-2\) and \(x=2\). At \(x=-2\), \(f'\) changes \(+\to-\): local maximum. At \(x=2\), \(f'\) changes \(-\to+\): \(\boxed{\text{local minimum at }x=2\text{, local maximum at }x=-2}\).
  2. 2(b) \(f'(x)>0\) (graph above the x-axis) for \(x<-2\) or \(x>2\), so \(\boxed{f\text{ is increasing on }x<-2\text{ or }x>2}\).
  3. 3(c) The \(f'\) graph is itself a parabola with its own turning point (vertex) at \(x=0\) — that is where the slope of \(f'\) is zero, i.e. \(f''(x)=0\). So \(\boxed{f\text{ has a point of inflection at }x=0}\).
  4. 4(d) The \(f'\) graph is increasing (rising left to right) for \(x>0\), so \(\boxed{f\text{ is concave up for }x>0}\) (and concave down for \(x<0\)).
The graph of f prime of x equals x squared minus 4An upward-opening parabola crossing the x-axis at x=-2 and x=2, with its own turning point (vertex) at (0,-4).−22(0,−4)y = f ′(x)
Remember: this is a picture of \(f'\), not \(f\) — every conclusion above comes from reading THIS curve's roots, sign, and turning point, never from solving an equation for \(f\).
Quick Check

The graph of \(f'(x)\) crosses the x-axis at \(x=5\), changing from positive to negative. What does this tell you about \(f\)?

To find the x-coordinate of a point of inflection of \(f\) from the graph of \(f'(x)\) alone, you should look for:

Rates of Change: Motion

If displacement is a function of time, its derivatives are velocity and acceleration.

From displacement to velocity to acceleration
\[v(t)=s'(t) \qquad a(t)=v'(t)=s''(t)\]
Reading the results
  • \(v(t)=0\): the object is momentarily at rest.
  • \(v(t)>0\): moving in the positive direction; \(v(t)<0\): moving in the negative direction.
  • \(a(t)=0\): velocity is momentarily not changing (often where speed is greatest or least).
Want more detail? Read Siyavula — 6.7 Applications of Differential Calculus
Worked example — velocity at a single instant Level 1

A particle moves in a straight line so that its displacement (in metres) after \(t\) seconds is \(s(t)=t^2-4t\). Determine its velocity at \(t=3\).

Show solution
  1. 1Velocity: \(v(t)=s'(t)=2t-4\).
  2. 2\(v(3)=2(3)-4=\boxed{2\text{ m/s}}\)
Worked example Level 3

A particle moves in a straight line so that its displacement (in metres) after \(t\) seconds is \(s(t)=t^3-6t^2+9t\). (a) Determine its velocity at \(t=2\). (b) Determine when the particle is at rest.

Show solution
  1. 1Velocity: \(v(t)=s'(t)=3t^2-12t+9\).
  2. 2(a) \(v(2)=3(4)-24+9=12-24+9=\boxed{-3\text{ m/s}}\) (moving in the negative direction).
  3. 3(b) At rest means \(v(t)=0\): \(3t^2-12t+9=0 \Rightarrow t^2-4t+3=0\).
  4. 4Factorise: \((t-1)(t-3)=0\), so \(\boxed{t=1\text{ s or }t=3\text{ s}}\).
The displacement-time graph of s(t)=t cubed minus 6t squared plus 9tA curve showing displacement rising, then falling to a local minimum near t=3, then rising again, with the particle momentarily at rest at t=1 and t=3.t=1t=3s(t)t
Reading the graph: the marked points at \(t=1\) and \(t=3\) are exactly where the tangent to the displacement curve is horizontal — velocity zero, the particle momentarily at rest, matching the calculus above.
Worked example — one instant, constant acceleration Level 2

A particle moves in a straight line so that its displacement (in metres) after \(t\) seconds is \(s(t)=t^2-6t+5\). Determine whether the particle is speeding up or slowing down at \(t=1\).

Show solution
  1. 1Velocity: \(v(t)=s'(t)=2t-6\). Acceleration: \(a(t)=v'(t)=2\) (a constant, always positive).
  2. 2\(v(1)=2(1)-6=-4<0\) and \(a(1)=2>0\).
  3. 3\(v(1)\) and \(a(1)\) have opposite signs, so \(\boxed{\text{the particle is slowing down at }t=1}\).
Worked example — speeding up or slowing down? Level 4

A particle moves in a straight line so that its displacement (in metres) after \(t\) seconds is \(s(t)=t^3-9t^2+24t\), for \(t\ge0\). Determine whether the particle is speeding up or slowing down (a) at \(t=1\), and (b) at \(t=5\).

Show solution
  1. 1Velocity: \(v(t)=s'(t)=3t^2-18t+24=3(t-2)(t-4)\). Acceleration: \(a(t)=v'(t)=6t-18=6(t-3)\).
  2. 2(a) At \(t=1\): \(v(1)=3(1-2)(1-4)=3(-1)(-3)=9>0\) and \(a(1)=6(1)-18=-12<0\).
  3. 3\(v(1)\) and \(a(1)\) have opposite signs (\(+\) and \(-\)) — acceleration is working against the direction of motion, so \(\boxed{\text{the particle is slowing down at }t=1}\).
  4. 4(b) At \(t=5\): \(v(5)=3(5-2)(5-4)=3(3)(1)=9>0\) and \(a(5)=6(5)-18=12>0\).
  5. 5\(v(5)\) and \(a(5)\) have the same sign (both \(+\)) — acceleration is reinforcing the direction of motion, so \(\boxed{\text{the particle is speeding up at }t=5}\).
The rule
A particle speeds up when \(v(t)\) and \(a(t)\) have the same sign (acceleration reinforces the motion) and slows down when they have opposite signs (acceleration opposes it) — a genuinely different question from just asking whether \(v(t)\) is positive or negative, and a common source of lost marks.
Sign diagram comparing velocity and acceleration to determine speeding up or slowing downTwo number lines for t from 0 to 6: the velocity sign row is positive before t=2, negative between t=2 and t=4, positive after t=4; the acceleration sign row is negative before t=3 and positive after t=3. A shaded band beneath combines both: amber where the signs differ, meaning slowing down, between t=0 and t=2, and between t=3 and t=4; green where the signs match, meaning speeding up, between t=2 and t=3, and between t=4 and t=6.v+−+a−+t=2t=3t=4
Reading the diagram: the shaded band combines the \(v\) and \(a\) sign rows above it — amber where they disagree (slowing down), green where they agree (speeding up). Note \(t=1\) falls in the first amber band and \(t=5\) falls in the last green band, matching the worked example.

Optimisation

Optimisation problems ask for the best (maximum or minimum) value of a quantity that depends on one variable — calculus finds it exactly.

Method
1

Write an expression for the quantity to be optimised, in terms of one variable only (use any given constraint to eliminate a second variable).

2

Differentiate the expression and set the derivative equal to zero.

3

Solve for the variable, checking it makes sense in context (e.g. a length can’t be negative).

4

Confirm it is a maximum or minimum with the second derivative, then answer the actual question asked.

Want more detail? Read Siyavula — 6.7 Applications of Differential Calculus
Worked example — pure algebra, no context Level 1

Determine the value of \(x\) that minimises \(f(x)=x^2-6x+13\), and this minimum value.

Show solution
  1. 1Differentiate: \(f'(x)=2x-6\).
  2. 2Set \(f'(x)=0\): \(2x-6=0 \Rightarrow x=3\).
  3. 3Confirm a minimum: \(f''(x)=2>0\) for every \(x\), so \(x=3\) is indeed a minimum.
  4. 4\(\boxed{f(3)=9-18+13=4}\)
Worked example — optimising a given formula Level 3

A company's monthly profit, in thousands of rand, from selling \(x\) units is \(P(x)=-3x^2+60x-180\) for \(0\le x\le20\). Determine the number of units that must be sold to maximise profit, and the maximum profit.

Show solution
  1. 1The function is already given — no geometry to set up, so differentiate directly: \(P'(x)=-6x+60\).
  2. 2Set \(P'(x)=0\): \(-6x+60=0 \Rightarrow x=10\), which lies inside the valid domain \(0\le x\le20\).
  3. 3Confirm a maximum: \(P''(x)=-6<0\) for every \(x\), so \(x=10\) is indeed a maximum.
  4. 4\(P(10)=-3(100)+600-180=-300+600-180=120\).
  5. 5\(\boxed{10\text{ units must be sold, for a maximum profit of R120\,000}}\).
Not every optimisation needs geometry
Sometimes the function to optimise is handed to you directly (a cost, profit or revenue formula) instead of being built from a shape's dimensions. This is the easier case — the harder version, where you have to build the formula yourself first, follows next.
Worked example — build the formula yourself, from geometry Level 4

An open-top box is made from a square sheet of cardboard with sides 12 cm, by cutting a square of side \(x\) cm from each corner and folding up the sides. (a) Show that the volume is \(V(x)=4x^3-48x^2+144x\). (b) Determine the value of \(x\) that maximises the volume, and this maximum volume.

Show solution
  1. 1(a) The base has side \((12-2x)\) and the height is \(x\), so \(V(x)=x(12-2x)^2=x(144-48x+4x^2)=\boxed{4x^3-48x^2+144x}\).
  2. 2(b) Differentiate: \(V'(x)=12x^2-96x+144=12(x^2-8x+12)=12(x-2)(x-6)\).
  3. 3Set \(V'(x)=0\): \(x=2\) or \(x=6\). Since \(0<x<6\) for the box to exist, only \(x=2\) is valid.
  4. 4Check it is a maximum: \(V''(x)=24x-96\), so \(V''(2)=48-96=-48<0\). Confirmed maximum.
  5. 5\(V(2)=4(8)-48(4)+144(2)=32-192+288=\boxed{128\text{ cm}^3\text{ at }x=2\text{ cm}}\).
The net of the box at x=2, a 12 by 12 square with a 2cm square cut from each cornerA flat square net 12cm by 12cm, with a 2cm by 2cm square removed from each of the four corners, leaving flaps that fold up along the inner dashed lines to form an open-top box with an 8cm by 8cm base and 2cm high sides.12 cm12 cmx=28 × 8 base
Reading the net: the four shaded \(2\times2\) corner squares are cut away; the remaining flaps fold up along the dashed lines to form the box, leaving an \(8\times8\) base (since \(12-2(2)=8\)) and walls of height \(x=2\) cm.
Quick Check

In an optimisation problem, after solving \(f'(x)=0\), why must you still check the second derivative (or a sign change)?

In the open-box problem on this page, why is \(x=6\) rejected even though \(V'(6)=0\)?

At a given instant, a particle's velocity is negative and its acceleration is positive. What is happening to its speed?

Core Formulas & Exam Strategy

Core formulas to know
\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\qquad \frac{d}{dx}\left[x^n\right]=nx^{n-1}\]\[f''(a)>0\Rightarrow\text{local minimum}\qquad f''(a)<0\Rightarrow\text{local maximum}\qquad f''(a)=0\Rightarrow\text{possible inflection}\]
Exam mantra
Differentiate. Solve. Interpret. Find the derivative, solve the equation the question actually asks for, then answer in the exact form and units the question wants — a common way to lose marks is stopping at the x-value when the question asked for a maximum volume, a y-coordinate, or an interval.
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What CAPS Expects You to Know

The Grade 12 Calculus knowledge statement this page is built from.

  1. 1

    Have an intuitive understanding of the limit concept, in the context of approximating the rate of change of a function at a point.

  2. 2

    Use limits to define the derivative of a function \(f\) at any \(x\), with the understanding that \(f'(a)\) is the gradient of the tangent to the graph of \(f\) at \(x=a\).

  3. 3

    Differentiate from first principles — examinable only on constant functions, quadratic-type polynomials \(ax^2+bx+c\), and reciprocal functions \(\dfrac{a}{x}\).

  4. 4

    Use the power rule together with the constant-multiple and sum/difference rules. No product or quotient rule is required — expand any product into separate terms first.

  5. 5

    Find equations of tangents to graphs of functions.

  6. 6

    Introduce the second derivative and how it determines the concavity of a function.

  7. 7

    Sketch graphs of cubic polynomial functions using differentiation to determine stationary points and points of inflection, and the factor theorem for x-intercepts, including a description of end behaviour as \(x\to\pm\infty\).

  8. 8

    Solve practical problems concerning optimisation and rate of change, including calculus of motion.

  • Calculus is examined in Term 2 of Grade 12 and is the single highest-weighted Paper 1 topic: 35±3 of 150 marks.
  • Average gradient and an intuitive sense of gradient-at-a-point were already introduced in Grade 11 — this page assumes that as revision, not new content.

How to Use This Lesson

A few practical notes before you start.

  • Work through the ideas in order — later slides (tangents, stationary points, sketching) all reuse the differentiation rules from earlier slides.
  • Several slides reuse the same function, \(f(x)=x^2-4x\), so you see one function answer several different types of question.
  • Try each worked example yourself before pressing “Show solution.”
  • Before you submit an answer, check: correct rule, careful substitution, and a final answer in the exact form the question asked for.

Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

First Principles

A worked walkthrough of differentiating using the limit definition.

Kevinmathscience · First Principles Calculus Grade 12

Rules of Differentiation

The power rule and basic differentiation rules, applied step by step.

Khan Academy · Power rule | Derivative rules

Sketching Cubic Graphs

Using stationary points and concavity to sketch a cubic with calculus.

NTE · How to Sketch a Cubic Function

Optimisation

A worked optimisation word problem, solved with derivatives.

Khan Academy · Minimizing combined area

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Not rewriting roots and reciprocals as powers first

The power rule only applies once every term is written as \(x^n\) — rewrite \(\sqrt{x}\) and \(1/x^n\) before differentiating, not after.

Finding a stationary point but never classifying it

\(f'(x)=0\) only locates the point — you must still use \(f''(x)\) or a sign change to say maximum or minimum.

Confusing \(f'(x)=0\) with \(f''(x)=0\)

\(f'(x)=0\) gives stationary points; \(f''(x)=0\) gives the point of inflection — they answer different questions.

Sign errors on subtracted or negative terms

Expand brackets like \(-(x+h)\) carefully — a dropped negative sign silently breaks the whole first-principles derivation.

Not rejecting values outside the valid domain

In optimisation, a stationary point that makes a length zero or negative is not a valid answer — check the physical constraints given.

Stopping too early

Finding \(x\) is often only step one — if the question asks for a maximum volume or a coordinate, substitute back and give that final value.

Reading a gradient straight off a general-form line

A line like \(4x-y+5=0\) must be rearranged into \(y=mx+c\) first — the gradient is \(m\), never one of the original equation's coefficients directly.

Looking for x-intercepts of f on a graph of f′

When only \(f'(x)\)'s graph is given, its roots are \(f\)'s turning points, not \(f\)'s x-intercepts — without an equation for \(f\), its own x-intercepts usually can't be found at all.

Related Topics

Study these alongside Calculus for a stronger Paper 1 foundation.

Functions & Graphs

Sketching and interpreting graphs

Study this

Algebra & Equations

Factorising and solving equations

Study this

Polynomials

Factor and remainder theorems

Study this

Practise This Topic

You've done the notes above — now practise and test yourself.

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Past Papers
Calculus Grade 12 Past Papers

Exam-style Grade 12 calculus questions arranged by cognitive level, with real citations from the DBE/provincial archive.

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Calculus Grade 12 Test Your Knowledge

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Frequently Asked Questions

Straight answers to common Grade 12 CAPS questions about Calculus.

What is new in Grade 12 Calculus?

Grade 12 introduces the formal, limit-based definition of the derivative, differentiation from first principles, the full differentiation toolkit, tangents, stationary points, concavity and points of inflection, sketching cubic graphs with calculus, and applying differentiation to rate-of-change and optimisation problems.

Is differentiation from first principles examined every year?

Yes, CAPS explicitly lists it as examinable, but only on a restricted set of function types: constant functions, quadratic-type polynomials of the form \(ax^2+bx+c\), and reciprocal functions of the form \(a/x\).

Do I need the product rule or quotient rule?

No. CAPS Grade 12 does not include a product rule or quotient rule. Expand any product into a sum of terms first, then differentiate term by term.

How do I sketch a cubic without a table of values?

Find the y-intercept by substituting \(x=0\), the x-intercepts using the factor theorem, the stationary points by solving \(f'(x)=0\), and the point of inflection by solving \(f''(x)=0\). Plot these points, then draw one smooth curve through them.

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Past Question Papers, then finish with the Test Your Knowledge quiz as a self-check.