Misclassifying Stationary Points

Q: Can I tell a maximum from a minimum just by the y-value?

No. A larger yyy does not make a point a maximum, the nature depends on how the curve behaves around it, not on its height. You must test either the sign of dydx\frac{dy}{dx}dxdy on both sides, or the sign of d2ydx2\frac{d^{2}y}{dx^{2}}dx2d2y at the point.

Q: What if the sign of the gradient is the same on both sides?

Then the stationary point is a point of inflection, not a turning point. For example, if dydx\frac{dy}{dx}dxdy is positive just before and just after the point, the curve keeps increasing through a momentary flat spot.

The error

Finding a stationary point is only half the job: $\frac{dy}{dx}=0$ locates it, but you still have to decide whether it is a maximum, a minimum, or a point of inflection. Three misclassifications are common.

The first is naming the nature straight from the $y$ -value, "this one is higher, so it must be the maximum", with no test at all. The second is skipping classification entirely and losing those marks.

The third, and most persistent, is reversing the first-derivative (sign-change) test: reading a slope that goes positive-then-negative as a minimum, when that rising-then-falling shape is actually a peak. It is tempting because the words "maximum" and "positive" feel linked, and because under time pressure the sign diagram gets filled in from memory rather than from the shape of the curve.

The result is a turning point labelled with the opposite nature, and the marks for determining that nature are lost even when the gradient work was perfect.

Wrong line, then the fix

For $y=x^{3}-3x^{2}+4$ , $\frac{dy}{dx}=3x^{2}-6x=3x(x-2)=0$ gives stationary points at $x=0$ and $x=2$ . Test the sign of $\frac{dy}{dx}$ on each side:

Interval	Test x	Sign of dy/dx
x < 0	x = -1	3(-1)(-3) = +9 (+)
0 < x < 2	x = 1	3(1)(-1) = -3 (−)
x > 2	x = 3	3(3)(1) = +9 (+)

Wrong, "at $x=0$ the slope goes $+$ then $-$ , so it is a minimum." That reverses the pattern.

Correct, at $x=0$ the slope is positive then negative: the curve rises then falls, so it is a maximum, at $(0,4)$ . At $x=2$ the slope is negative then positive: the curve falls then rises, so it is a minimum, at $(2,0)$ .

The reliable fix

1
Find every stationary point
Solve $\frac{dy}{dx}=0$ for all its $x$ -values. Classification only applies to these points.
2
Choose test points around each
Pick one $x$ just below and one just above each stationary point, without crossing a neighbouring stationary point.
3
Read the sign, not the size
Substitute into $\frac{dy}{dx}$ and record only whether it is positive or negative on each side.
4
Match the shape to the nature
Rising then falling ( $+$ then $-$ ) is a maximum; falling then rising ( $-$ then $+$ ) is a minimum; same sign both sides is a point of inflection.
5
State the point in full
Substitute the stationary $x$ back into $y$ to give the coordinates, then name the nature.

One line to remember it

Up-then-down is a peak, down-then-up is a valley

Slope positive then negative means the curve climbed and then fell, a peak, so a maximum. Slope negative then positive means it dipped and then rose, a valley, so a minimum.

Practice where the error hides

Q1[6 marks]

Find the stationary points of the curve $y=2x^{3}-3x^{2}-12x+5$ and determine the nature of each using the first-derivative test.

Show worked solution

Differentiate and set the result to zero:

\frac{dy}{dx}=6x^{2}-6x-12=6(x^{2}-x-2)=6(x-2)(x+1)=0

So the stationary points are at $x=-1$ and $x=2$ . Test the sign of $\frac{dy}{dx}=6(x-2)(x+1)$ on each side:

Test x	6(x-2)(x+1)	Sign
x = -2	6(-4)(-1) = +24	(+)
x = 0	6(-2)(1) = -12	(−)
x = 3	6(1)(4) = +24	(+)

At $x=-1$ the slope goes $+$ then $-$ : rising then falling, so a maximum. Its $y$ -value is $y=2(-1)^{3}-3(-1)^{2}-12(-1)+5=-2-3+12+5=12$ , giving the maximum point $(-1,12)$ .

At $x=2$ the slope goes $-$ then $+$ : falling then rising, so a minimum. Its $y$ -value is $y=2(2)^{3}-3(2)^{2}-12(2)+5=16-12-24+5=-15$ , giving the minimum point $(2,-15)$ .

So $(-1,12)$ is a maximum and $(2,-15)$ is a minimum. Judging only by the $y$ -values would have accidentally agreed here, but the sign test is what proves it, and it is the method that survives when a curve has just one turning point.

How one-to-one teaching helps

Our teachers replace guessing with a locked routine: solve $\frac{dy}{dx}=0$ , test the sign on each side, and read the shape aloud, "up then down, peak", before writing the word maximum or minimum. In a one-to-one lesson we catch the exact moment the pattern flips and re-anchor it with a quick sketch of a hill and a valley, so the picture, not the memory, decides the nature.

Because the paper awards method marks, a clear sign diagram earns credit even if a later value slips. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To drill classification cleanly, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Can I tell a maximum from a minimum just by the y-value?

No. A larger $y$ does not make a point a maximum, the nature depends on how the curve behaves around it, not on its height.

You must test either the sign of $\frac{dy}{dx}$ on both sides, or the sign of $\frac{d^{2}y}{dx^{2}}$ at the point.

Which slope pattern is a maximum?

Positive then negative. The curve rises up to the stationary point and then falls away, a peak, so it is a maximum.

Negative then positive is a valley and gives a minimum.

What if the sign of the gradient is the same on both sides?

Then the stationary point is a point of inflection, not a turning point. For example, if $\frac{dy}{dx}$ is positive just before and just after the point, the curve keeps increasing through a momentary flat spot.

Is the first-derivative test or the second-derivative test better?

Both are valid. The first-derivative (sign-change) test always works, even when the second derivative is zero.

The second-derivative test is faster when it applies. Choose whichever you can carry out without reversing the convention.

Source:SRC-DSKP-EN

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

Misclassifying stationary points

The error

Wrong line, then the fix

The reliable fix

One line to remember it

Practice where the error hides

How one-to-one teaching helps

Frequently asked questions

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