What are Stationary Points?

stationary points
Stationary points (or turning/critical points) are the points on a curve where the gradient is 0. This means that at these points the curve is flat. Usually, the gradient of a curve is always changing and so the gradient is only 0 instantaneously (unless the curve is a flat line, in which case, the gradient is always 0).

A MAXIMUM is located at the top of a peak on a curve. Conversely, a MINIMUM if it is at the bottom of a trough.

A stationary point can be found by solving \frac{dy}{dx}=0, i.e. finding the x coordinate where the gradient is 0. See more on differentiating to find out how to find a derivative.

See Example 1.

Click here for an online tool for checking your stationary points. Page 21 onwards of the StudyWell Differentiation eGuide has more on Stationary Points including exam-style questions.

Classifying Stationary Points

For certain functions, it is possible to differentiate twice (or even more) and find the second derivative. It is often denoted as f''(x) or \frac{d^2y}{dx^2}. For example, given that f(x)=x^7-x^5 then the derivative is f'(x)=7x^6-5x^4 and the second derivative is given by f''(x)=42x^5-20x^3.

The second derivative can tell us something about the nature of a stationary point:

  • For a MINIMUM, the gradient changes from negative to 0 to positive, i.e. the gradient is increasing. Hence, the second derivative is positive – f''(x)> 0.
  • For a MAXIMUM, the gradient changes from positive to 0 to negative, i.e. the gradient is decreasing. Hence, the second derivative is negative – f''(x)< 0.

We can classify whether a point is a minimum or maximum by determining whether the second derivative is positive or negative. This is done by putting the x-coordinates of the stationary points into f''(x).

Page 21 onwards of the StudyWell Differentiation eGuide has more on Stationary Points including exam-style questions.


Find the stationary points on the curve y=\frac{2}{3}x^3-5x^2+8x-4.


Start by solving \frac{dy}{dx}=0:

\frac{dy}{dx}=2x^2-10x+8=0 i.e. x^2-5x+4=0. Factorising gives (x-4)(x-1)=0 and so the x coordinates are x=4 and x=1. Substituting these into the y equation gives the coordinates of the turning points as (4,-28/3) and (1,-1/3).

Find and classify the stationary points of f(x)=x^3-2x^2+x-5.


We first locate them by solving f'(x)=0. f'(x) is given by


We can solve f'(x)=0 by factorising:

which gives x=1/3 or x=1. The corresponding y coordinates are \left(\frac{1}{3}\right)^3-2\left(\frac{1}{3}\right)^2+\frac{1}{3}-5=-\frac{131}{27} (don’t be afraid of strange fractions) and (1)^3-2(1)^2+1-5=-5. Hence, the critical points are at (1/3,-131/27) and (1,-5). We can classify them by substituting the x coordinate into the second derivative and seeing if it is positive or negative. Differentiating a second time gives
f''(x)=6x-4. It follows that f''(1/3)=6\times\frac{1}{3}-4=-2 which is less than 0, and hence (1/3,-131/27) is a MAXIMUM. Similarly, f''(1)=6times 1-4=2> 0 and (1,-5) is a MINIMUM.



Using differentiation to locate and classify the minimum of the cost of a journey.


Using stationary points to sketch a functions that is a combination of a reciprocal and a cubic function.