Skip to content
spmaddmath.com.my
Tuition

Study

SyllabusFormulasMethodsExam & PapersTools
LocationsPricingBlogOur TeachersContact
EN

Calculus Clinic · Differentiation

Misusing the second-derivative test

At a stationary point, a positive second derivative means a minimum and a negative one means a maximum. Get the sign convention backwards, or apply the test where dydx0\frac{dy}{dx}\neq 0, and the nature comes out wrong.

The error

The second-derivative test decides whether a stationary point is a maximum or a minimum. The rule is: at a point where dydx=0\frac{dy}{dx}=0, if d2ydx2<0\frac{d^{2}y}{dx^{2}}<0 the point is a maximum, and if d2ydx2>0\frac{d^{2}y}{dx^{2}}>0 it is a minimum.

Two misuses are common. The first is reversing the convention, reading a positive second derivative as a maximum, because 'positive' feels like 'the top'.

The second is applying the test at a point that is not stationary at all, so the sign of d2ydx2\frac{d^{2}y}{dx^{2}} there decides nothing about a turning point. Both are tempting under time pressure, when the routine gets run from memory without checking which value goes with which conclusion.

Wrong line, then the fix

For y=x26x+5y=x^{2}-6x+5, dydx=2x6=0\frac{dy}{dx}=2x-6=0 gives the stationary point at x=3x=3. The second derivative there is:

d2ydx2=2>0\frac{d^{2}y}{dx^{2}}=2>0

The reversed convention draws the wrong conclusion: Wrong,'the second derivative is positive, so the point is a maximum.'

Correct, a positive second derivative means the curve is concave up, shaped like a valley, so the stationary point at x=3x=3 is a minimum. A negative second derivative would mean concave down, a hill, and hence a maximum.

The reliable fix

  1. 1

    Find the stationary points first

    Solve dydx=0\frac{dy}{dx}=0. The test only decides the nature of these points, nowhere else.

  2. 2

    Differentiate again

    Find d2ydx2\frac{d^{2}y}{dx^{2}} as a function of xx.

  3. 3

    Substitute the stationary x-value

    Put each stationary xx into d2ydx2\frac{d^{2}y}{dx^{2}} and read its sign.

  4. 4

    Apply the convention

    d2ydx2>0\frac{d^{2}y}{dx^{2}}>0\Rightarrow minimum; d2ydx2<0\frac{d^{2}y}{dx^{2}}<0\Rightarrow maximum.

  5. 5

    Handle the zero case

    If d2ydx2=0\frac{d^{2}y}{dx^{2}}=0, the test is inconclusive; check the sign of dydx\frac{dy}{dx} just before and just after the point instead.

One line to remember it

Positive smiles, negative frowns

A positive second derivative is concave up, a smile, so it is a minimum. A negative second derivative frowns, so it is a maximum.

Practice where the error hides

Q1[5 marks]

Find the stationary points of the curve y=x312x+5y=x^{3}-12x+5 and determine the nature of each using the second-derivative test.

Show worked solution

Differentiate and set the result to zero to find the stationary points.

dydx=3x212=0    x2=4    x=±2\frac{dy}{dx}=3x^{2}-12=0\;\Rightarrow\; x^{2}=4\;\Rightarrow\; x=\pm 2

Find the second derivative:

d2ydx2=6x\frac{d^{2}y}{dx^{2}}=6x

At x=2x=2: d2ydx2=6(2)=12>0\frac{d^{2}y}{dx^{2}}=6(2)=12>0, so this point is a minimum. Its yy-value is y=(2)312(2)+5=824+5=11y=(2)^{3}-12(2)+5=8-24+5=-11, giving the minimum point (2,11)(2,-11).

At x=2x=-2: d2ydx2=6(2)=12<0\frac{d^{2}y}{dx^{2}}=6(-2)=-12<0, so this point is a maximum. Its yy-value is y=(2)312(2)+5=8+24+5=21y=(-2)^{3}-12(-2)+5=-8+24+5=21, giving the maximum point (2,21)(-2,21).

So (2,11)(2,-11) is a minimum and (2,21)(-2,21) is a maximum. The positive second derivative belongs to the minimum and the negative one to the maximum, the convention applied the right way round.

How one-to-one teaching helps

The fix is not more theory but a locked routine, and that is what our teachers build. We drill 'stationary first, sign second': solve dydx=0\frac{dy}{dx}=0, then read the sign of d2ydx2\frac{d^{2}y}{dx^{2}} only at those xx-values, and say 'positive smiles, minimum' aloud each time until it sticks.

In a one-to-one lesson we can catch the exact moment the convention flips and re-anchor it with a quick sketch of a valley and a hill. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

Does a positive second derivative mean maximum or minimum?

Minimum. A positive second derivative means the curve is concave up, shaped like a valley, so a stationary point there is a minimum.

A negative second derivative means concave down and gives a maximum.

Can I use the test at any point on the curve?

No. The second-derivative test only classifies stationary points, where dydx=0\frac{dy}{dx}=0.

The sign of d2ydx2\frac{d^{2}y}{dx^{2}} at a non-stationary point tells you about concavity, not about a maximum or minimum.

What if the second derivative is zero at a stationary point?

Then the test is inconclusive, the point could be a maximum, a minimum, or a point of inflection. Fall back on the first-derivative test: check the sign of dydx\frac{dy}{dx} just before and just after the stationary point.

Is the second-derivative test worth using if I might reverse it?

Yes, it is fast and clean once the convention is secure. Anchor it with 'positive smiles (minimum), negative frowns (maximum)', write that reminder on your working, and the reversal stops happening.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Ready to get started?

Book a Trial Classfrom RM50/hr · One-hour paid trial · Same-day reply
Book a Trial ClassOne-hour paid trial · Same-day reply