## 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.

## Finding Stationary Points

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.

### Example

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).

## 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 $f$ or $\frac{d^2y}{dx^2}$. For example, given that $f(x)=x^7-x^5$ then the derivative is $f$ and the second derivative is given by $f$.

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 – $f$.
• For a MAXIMUM, the gradient changes from positive to 0 to negative, i.e. the gradient is decreasing. Hence, the second derivative is negative – $f$.

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 $f$.

### Example

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

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

$f$

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

$(3x-1)(x-1)=0$
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
$f$. It follows that $f$ which is less than 0, and hence (1/3,-131/27) is a MAXIMUM. Similarly, $f$ and (1,-5) is a MINIMUM.

Exam questions that find and classify stationary points quite often have a practical context. Click here to find Questions by Topic and scroll down to all past DIFFERENTIATION – OPTIMISATION questions to practice this type of question.

Are you ready to test your Pure Maths knowledge? If so, visit our Practice Papers page and take StudyWell’s own Pure Maths tests.