## 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 , 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 .*

Start by solving :

i.e. . Factorising gives 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).

Click here for an online tool for checking your stationary points.

**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 or . For example, given that then the derivative is and the second derivative is given by .

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

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 -coordinates of the stationary points into .

**Example**

*Find and classify the stationary points of .
*

We first locate them by solving . 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 (don’t be afraid of strange fractions) and . 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

. It follows that which is less than 0, and hence (1/3,-131/27) is a MAXIMUM. Similarly, 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.

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