## 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.

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