# Stationary Points (turning points) – maxima & minima

## What are 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.

## Examples

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

$f'(x)=3x^2-4x+1$.

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

$f”(x)=6x-4$. 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, $f”(1)=6\times 1-4=2> 0$ and $(1,-5)$ is a MINIMUM.

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