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What does the second derivative tell us about the shape of a curve and the nature of its stationary points?

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

WACE Year 12 Mathematics Methods Unit 3 the second derivative: concavity, points of inflection where the second derivative changes sign, the second derivative test for stationary points, and worked SCSA-style examples.

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  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

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 ff', 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.

A common requirement is that f(a)=0f''(a)=0 alone does not guarantee an inflection: the sign of ff'' 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 ff'' 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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksCalculator-free. Consider f(x)=x33x29xf(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)=3x26x9=3(x3)(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)=6x6f''(x)=6x-6. At x=3x=3, f(3)=12>0f''(3)=12>0, a local minimum, with f(3)=272727=27f(3)=27-27-27=-27. At x=1x=-1, f(1)=12<0f''(-1)=-12<0, a local maximum, with f(1)=13+9=5f(-1)=-1-3+9=5.

(b) f(x)=0f''(x)=0 at x=1x=1, where ff'' changes sign. f(1)=139=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 20244 marksCalculator-free. The displacement of a particle is x(t)=t36t2+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.
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A kinematics application of the second derivative.

(a) v(t)=x(t)=3t212t+12v(t)=x'(t)=3t^{2}-12t+12 and a(t)=v(t)=x(t)=6t12a(t)=v'(t)=x''(t)=6t-12.

(b) a(t)=0a(t)=0 when 6t12=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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