Max & Min Points (Turning Points) Revision Notes for Leaving Cert Mathematics

Max & Min Points (Turning Points)

Introduction

For most graphs, you may notice points for which the graph begins to "turn". That is, the graph goes from strictly increasing to strictly decreasing (or vice versa). These points are known as stationary points (or turning points).

In the following diagram, the turning points are $(0,0)$ and $(1,1)$ . At these points exactly, the slope of the tangents is neither increasing nor decreasing. The slope is actually completely flat.

Example

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Determine the turning points of the function $f(x)=3x^2-2x^3$ .

First differentiate the function :

f'(x)=6x-6x^2

The turning points occur where the slopes of the tangents are flat, so the slope has to be $0$ .

\begin{align*} 6x-6x^2&=0 \\\\ x-x^2&= \\\\ x(1-x) &= \\\\ x=0, x&=1 \end{align*}

We found the $x$ coordinates of the turning points, to determine the $y$ coordinates, substitute back into the original function.

f(0)=3(0)^2-2(0)^3=0 \\ f(1)=3(1)^2-2(1)^3=1

So, the turning points are :

(0,0), (1,1)

Second Derivative Test

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In the previous example we've determined the turning points of the function, but this doesn't tell us which is the local minima and which one is the local maxima. To find this out, we need to do the second derivative test. Take the second derivative of the function $f$ :

f''(x)=6-12x

If $x_1$ is a turning point, $x_1$ is a local minima if $f''(x_1)>0$ .

If $x_1$ is a turning point, $x_1$ is a local maxima if $f''(x_2)<0$ .

Let's test the point $x=0$ :

f''(0)=6-12(0)=6

$6>0$ , so the point $(0,0)$ is a local minima.

Now test the point $x=1$

f''(1)=6-12(1)=-6

$-6<0$ , so the point $(1,1)$ is a local maxima.

Max & Min Points (Turning Points) (Leaving Cert Mathematics): Revision Notes