Aus Online Mathematik Brückenkurs 2

There are three types of points at which the function can have local extreme points:

1. Critical points, i.e. where \displaystyle {f}'\left( x \right)=0;

2. Points where the function is not differentiable;

3. Endpoints of the interval of definition.

Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy \displaystyle 2 and \displaystyle 3.

As regards \displaystyle 1, we set the derivative equal to zero and obtain the equation

\displaystyle {f}'\left( x \right)=6x^{2}+6x-12=0

Dividing both sides by \displaystyle 6 and completing the square, we obtain

\displaystyle \left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0

This gives us the equation

\displaystyle \left( x+\frac{1}{2} \right)^{2}=\frac{9}{4}

and taking the root gives the solutions

\displaystyle x=-\frac{1}{2}-\sqrt{\frac{9}{4}}=-\frac{1}{2}-\frac{3}{2}=-2

\displaystyle x=-\frac{1}{2}+\sqrt{\frac{9}{4}}=-\frac{1}{2}+\frac{3}{2}=1

This means that if the function has several extreme points, they must lie between \displaystyle x=-2\text{ } and \displaystyle x=1.

Then, we write down a sign table for the derivative, and read off the possible extreme points.

TABLE

The function has a local maximum at \displaystyle x=-2\text{ } and a local minimum at \displaystyle x=1.

We obtain the overall appearance of the graph of the function from the table and by calculating the value of the function at a few points.

PICTURE TABLE