Processing Math: Done

Aus Online Mathematik Brückenkurs 2

We need to investigate three types of points in order to determine the function's local extreme points,

1. Critical points, i.e. where fx=0 ;

2. Points where the function is not differentiable;

3. Endpoints of the interval of definition.

but we can ignore 2. and 3., because the function is a polynomial and is therefore defined and differentiable everywhere.

The critical points are given by the points where the derivative,

fx=−4x3+83x2−182x=−4x3+24x2−36x=−4xx2−6x+9

is equal to zero.

From the factorized from of the derivative, we see that the derivative is zero when either the first factor, x, is zero, or when the other factor is zero:

x2−6x+9=0

We solve this second-degree equation by completing the square on the left-hand side,

x−32−32+9=0 

which are, after simplifying

x−32=0 

and this equation has the root x=3.

Therefore, the function has two critical points, x=0 and x=3.

The next step is to write down an outline of the derivative's sign changes, from which we then can see the possible local extreme points.

Because the derivative can be written as

fx=−4xx−32 

we start by writing down the sign changes for the factors −4x and x−32 .

TABLE

The derivative, which is the product of these factors, has the sign changes given below, which are a consequence of the calculating rules for signs:

++=+−+=−and−−=+

TABLE

From this, we see that x=0 is a local maximum, whilst x=3 is an inflexion point (and therefore not an extreme point).