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The second derivative and concavity: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Find and interpret the second derivative, determine concavity and points of inflection, and apply the second derivative test”

8 min read

Quick answer

The second derivative f′′(x)f''(x) measures concavity: f′′>0f''>0 is concave up, f′′<0f''<0 is concave down. A point of inflection is where f′′f'' changes sign. The second derivative test classifies a stationary point as a minimum if f′′>0f''>0 and a maximum if f′′<0f''<0.

Jump to a section
  1. What this dot point is asking
  2. What the second derivative measures
  3. Points of inflection
  4. The second derivative test
  5. Acceleration and the second derivative
  6. Reading concavity from a sketch
  7. Exam-style questions

What this dot point is asking

SCSA Unit 3 introduces the second derivative as the derivative of the derivative. This dot point asks you to compute it, read concavity from its sign, locate points of inflection, and use it to classify stationary points, all examinable in both sections of the WACE written examination.

What the second derivative measures

The first derivative f′(x)f'(x) is the rate of change of ff. The second derivative f′′(x)f''(x) is the rate of change of f′f', so it measures how the gradient is changing. In kinematics, if x(t)x(t) is displacement then x′(t)x'(t) is velocity and x′′(t)x''(t) is acceleration.

Concavity from the second derivative
  • f′′(x)>0f''(x)>0: the gradient is increasing, so the curve is concave up.
  • f′′(x)<0f''(x)<0: the gradient is decreasing, so the curve is concave down.
  • A point of inflection occurs where f′′(x)f''(x) changes sign (the concavity reverses).

A common requirement is that f′′(a)=0f''(a)=0 alone does not guarantee an inflection: the sign of f′′f'' must actually change there. Confirming the sign change, with a brief sign table or a sentence stating the signs on each side, is the part SCSA markers look for and the part most often omitted.

Points of inflection

To find points of inflection, solve f′′(x)=0f''(x)=0, then confirm a sign change of f′′f'' on either side. The function f(x)=x4f(x)=x^{4} illustrates the trap: f′′(x)=12x2f''(x)=12x^{2} is zero at x=0x=0 but never negative, so the concavity does not reverse and there is no inflection.

Locating a point of inflection

Step 1: Differentiate twice to obtain f′′(x)f''(x)

Differentiating f(x)=x3−6x2+5f(x)=x^{3}-6x^{2}+5 twice gives the second derivative that will reveal where concavity may change:

f′(x)=3x2−12x,f′′(x)=6x−12.f'(x)=3x^{2}-12x, \qquad f''(x)=6x-12.

Step 2: Solve f′′(x)=0f''(x)=0 to find the candidate

Setting f′′(x)=0f''(x)=0 locates where the concavity could reverse; this is a necessary condition, not a sufficient one:

6x−12=0  ⇒  x=2.6x-12=0 \;\Rightarrow\; x=2.

Step 3: Confirm the sign change of f′′f'' on either side

For x<2x<2, f′′<0f''<0 (concave down); for x>2x>2, f′′>0f''>0 (concave up). Because the concavity reverses, x=2x=2 is a genuine point of inflection.

Step 4: Find the yy-coordinate

Substituting x=2x=2 into the original function gives the yy-value of the inflection point:

f(2)=8−24+5=−11.f(2)=8-24+5=-11.

Final answer: Point of inflection at (2,−11)(2,-11)

The second derivative test

Once a stationary point x=ax=a is found from f′(a)=0f'(a)=0, the second derivative test classifies it quickly.

The second derivative test

At a stationary point where f′(a)=0f'(a)=0:

  • if f′′(a)>0f''(a)>0, then x=ax=a is a local minimum;
  • if f′′(a)<0f''(a)<0, then x=ax=a is a local maximum;
  • if f′′(a)=0f''(a)=0, the test is inconclusive; use a sign table for f′f'.
Classifying with the second derivative test

Step 1: Find the stationary points from f′(x)=0f'(x)=0

Differentiating f(x)=x3−3xf(x)=x^{3}-3x and setting the result to zero locates the candidates for classification:

f′(x)=3x2−3=0  ⇒  x=±1.f'(x)=3x^{2}-3=0 \;\Rightarrow\; x=\pm 1.

Step 2: Compute f′′(x)f''(x) and evaluate at each stationary point

The second derivative f′′(x)=6xf''(x)=6x is straightforward to evaluate. Its sign at each stationary point determines the nature of that point without needing a sign table for f′f':

  • At x=1x=1: f′′(1)=6>0f''(1)=6>0, so x=1x=1 is a local minimum.
  • At x=−1x=-1: f′′(−1)=−6<0f''(-1)=-6<0, so x=−1x=-1 is a local maximum.

Final answer: Local minimum at x=1x=1; local maximum at x=−1x=-1.

Acceleration and the second derivative

In kinematics the second derivative has a direct physical meaning. If x(t)x(t) is displacement, then x′(t)=v(t)x'(t)=v(t) is velocity and x′′(t)=a(t)x''(t)=a(t) is acceleration. Concave-up displacement (x′′>0x''>0) means the velocity is increasing, and concave-down displacement (x′′<0x''<0) means it is decreasing. A point of inflection of the displacement graph is exactly where the acceleration is zero and changing sign, which is the instant the motion switches between speeding up and slowing down. This connects the abstract idea of concavity to a quantity students can feel.

Reading concavity from a sketch

When you sketch a curve, concavity tells you how to draw it between the key points. A concave-up section curves like a valley with the gradient steadily increasing; a concave-down section curves like a hill with the gradient steadily decreasing. At a point of inflection the tangent crosses the curve as the bend reverses. Getting the concavity right is a marked feature of a curve sketch, separate from plotting the correct stationary points.

Common errors
Assuming f′′(a)=0f''(a)=0 means an inflection
You must confirm f′′f'' changes sign. f(x)=x4f(x)=x^{4} has f′′(0)=0f''(0)=0 but no inflection at x=0x=0.
Using the test when f′′(a)=0f''(a)=0
The second derivative test is inconclusive then; fall back on the sign of f′f' either side.
Reversing the concavity labels
Concave up corresponds to f′′>0f''>0 (a valley shape), concave down to f′′<0f''<0 (a hill shape).

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks
Calculator-free. Consider f(x)=x3−3x2−9xf(x)=x^{3}-3x^{2}-9x. (a) Find the stationary points and classify them using the second derivative test. (b) Find the coordinates of the point of inflection.
Show worked answer →

A standard stationary-point and inflection question.

(a) f′(x)=3x2−6x−9=3(x−3)(x+1)f'(x)=3x^{2}-6x-9=3(x-3)(x+1), so stationary points at x=3x=3 and x=−1x=-1. f′′(x)=6x−6f''(x)=6x-6. At x=3x=3, f′′(3)=12>0f''(3)=12>0, a local minimum, with f(3)=27−27−27=−27f(3)=27-27-27=-27. At x=−1x=-1, f′′(−1)=−12<0f''(-1)=-12<0, a local maximum, with f(−1)=−1−3+9=5f(-1)=-1-3+9=5.

(b) f′′(x)=0f''(x)=0 at x=1x=1, where f′′f'' changes sign. f(1)=1−3−9=−11f(1)=1-3-9=-11, so the inflection is (1,−11)(1,-11).

Markers reward both stationary points with the second-derivative test and the inflection with a sign-change justification.

WACE 2024-style4 marks
Calculator-free. The displacement of a particle is x(t)=t3−6t2+12tx(t)=t^{3}-6t^{2}+12t metres. (a) Find the acceleration a(t)a(t). (b) Find the time at which the acceleration is zero and describe the motion at that instant.
Show worked answer →

A kinematics application of the second derivative.

(a) v(t)=x′(t)=3t2−12t+12v(t)=x'(t)=3t^{2}-12t+12 and a(t)=v′(t)=x′′(t)=6t−12a(t)=v'(t)=x''(t)=6t-12.

(b) a(t)=0a(t)=0 when 6t−12=06t-12=0, so t=2t=2 s. At t=2t=2 the acceleration is zero, so the velocity is momentarily neither increasing nor decreasing; this is the point of inflection of the displacement graph, where the particle's speed is changing least.

Markers reward the second derivative as acceleration, t=2t=2, and the interpretation as zero acceleration.

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