Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.

jsMath

Lösung 1.3:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
K (Solution 1.3:2c moved to Lösung 1.3:2c: Robot: moved page)

Version vom 10:10, 11. Mär. 2009

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

  1. critical points, i.e. where f(x)=0,
  2. points where the function is not differentiable, and
  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 items 2 and 3.

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

f(x)=6x2+6x12=0.

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

x+2122122=0. 

This gives us the equation

x+212=49 

and taking the square root gives the solutions

xx=2149=2123=2=21+49=21+23=1.

This means that if the function has several extreme points, they must be among x=2 and x=1.

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

x 2 1
f(x) + 0 0 +
f(x) 21 6


The function has a local maximum at x=2 and a local minimum at 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.