Mathematics/Lessons/Differentiation/Stationary Points
Pure · Differentiation · Year 1

Stationary Points

Peaks and valleys sit where the gradient is zero. Solve f′(x)=0 to find them, then use f″(x) to tell which is which.

6 min read 1 worked example

Where the gradient is zero

A stationary point is a point on a curve where the gradient is zero — the curve is momentarily flat. These are the peaks and valleys: the maximum and minimum points. Because the gradient is $f'(x)$, finding them is simple in principle: solve $f'(x) = 0$. The interesting question is then what kind of stationary point each one is.

A curve labelled y = f(x) which rises to a local maximum, falls to a local minimum, then rises again. A dot marks each turning point, and through each dot a short horizontal line shows that the tangent there is flat. Both points are labelled f dash of x equals zero.
At a local maximum the curve stops rising and turns to fall; at a local minimum it stops falling and turns to rise. The tangent is horizontal at both — which is exactly why solving f′(x) = 0 finds them.
Why f′(x) = 0 locates peaks and valleys

Approaching a maximum, the curve is rising ($f'>0$); just past it, the curve is falling ($f'<0$). Right at the top it is neither — the gradient passes through zero. The same holds at a minimum, with the signs reversed. So every peak and valley sits where $f'(x)=0$. Solving that equation finds all the candidates in one step; classifying them tells you which are peaks and which are valleys.

Classifying with the second derivative

Once you have a stationary point, the second derivative tells you its nature at a glance:

$f''(x) < 0$  — the curve is concave (bending down): a maximum.
$f''(x) > 0$  — the curve is convex (bending up): a minimum.
$f''(x) = 0$  — inconclusive; check the gradient's sign either side instead.

Worked example 1

Find and classify stationary points
Find the stationary points of $y = x^3 - 3x^2 - 9x + 5$ and determine their nature.
MethodSolve f′=0, then test each with f″
1
Differentiate: $\dfrac{dy}{dx} = 3x^2 - 6x - 9$.
Power rule.
2
Set $=0$: $3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1)=0$.
Solve the quadratic.
3
So $x = 3$ or $x = -1$. The $y$-values: $y(3) = -22$ and $y(-1) = 10$.
Substitute into the curve.
4
Second derivative: $\dfrac{d^2y}{dx^2} = 6x - 6$.
Differentiate again.
5
At $x=3$: $6(3)-6 = 12 > 0$ ⇒ minimum $(3, -22)$. At $x=-1$: $6(-1)-6 = -12 < 0$ ⇒ maximum $(-1, 10)$.
Sign of $f''$ gives the nature.
▪ solved
Examiner's eye

Always give stationary points as full coordinates $(x, y)$ — the $y$ comes from the original curve, not the derivative. Then state the nature explicitly using the second derivative's sign: positive means minimum, negative means maximum. Doing both the find and the classify is what earns full marks.

Common mistake

Two traps catch most students. First, forgetting to classify: finding $f'(x)=0$ locates a stationary point but doesn't say whether it's a max or min — always test with $f''$. Second, giving only the $x$-value: a stationary point needs its $y$-coordinate too, read off the original curve, not the derivative.

Check yourself

Find the stationary point of $y = x^2 - 6x + 5$ and determine its nature using the second derivative.
(Hint: solve $\tfrac{dy}{dx}=0$ for $x$, find $y$ from the curve, then check the sign of $\tfrac{d^2y}{dx^2}$.)

Method
1
$\dfrac{dy}{dx} = 2x - 6$. Set $=0$: $2x - 6 = 0 \Rightarrow x = 3$.
Solve for x.
2
$y$-value: $y = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$. Point $(3, -4)$.
Substitute into the curve.
3
$\dfrac{d^2y}{dx^2} = 2 > 0$ for all $x$.
Second derivative is constant.
4
Since $f''>0$, the point $(3, -4)$ is a minimum.
Positive ⇒ minimum.
▪ solved

This is the vertex of an upward parabola, so a minimum is exactly what we expect — and $\tfrac{d^2y}{dx^2}=2>0$ confirms it. The turning point is $(3, -4)$.

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