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Lösung 1.3:2c

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.3:2c moved to Solution 1.3:2c: Robot: moved page)
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There are three types of points at which the function can have local extreme points:
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<center> [[Image:1_3_2c-1(3).gif]] </center>
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1. Critical points, i.e. where
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<math>{f}'\left( x \right)=0</math>;
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<center> [[Image:1_3_2c-2(3).gif]] </center>
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2. Points where the function is not differentiable;
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<center> [[Image:1_3_2c-3(3).gif]] </center>
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3. Endpoints of the interval of definition.
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Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy
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<math>2</math>
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and
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<math>3</math>.
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As regards
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<math>1</math>, we set the derivative equal to zero and obtain the equation
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<math>{f}'\left( x \right)=6x^{2}+6x-12=0</math>
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Dividing both sides by
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<math>6</math>
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and completing the square, we obtain
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<math>\left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0</math>
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This gives us the equation
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<math>\left( x+\frac{1}{2} \right)^{2}=\frac{9}{4}</math>
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and taking the root gives the solutions
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<math>x=-\frac{1}{2}-\sqrt{\frac{9}{4}}=-\frac{1}{2}-\frac{3}{2}=-2</math>
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<math>x=-\frac{1}{2}+\sqrt{\frac{9}{4}}=-\frac{1}{2}+\frac{3}{2}=1</math>
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This means that if the function has several extreme points, they must lie between
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<math>x=-2\text{ }</math>
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and
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<math>x=1</math>.
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Then, we write down a sign table for the derivative, and read off the possible extreme points.
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TABLE
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The function has a local maximum at
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<math>x=-2\text{ }</math>
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and a local minimum at
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<math>x=1</math>.
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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.
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PICTURE TABLE
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[[Image:1_3_2_c.gif|center]]
[[Image:1_3_2_c.gif|center]]

Version vom 11:01, 15. Okt. 2008

There are three types of points at which the function can have 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.

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

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


fx=6x2+6x−12=0 

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


x+212−212−2=0 

This gives us the equation


x+212=49 


and taking the root gives the solutions


x=−21−49=−21−23=−2 


x=−21+49=−21+23=1 


This means that if the function has several extreme points, they must lie between x=−2 and 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 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.

PICTURE TABLE


Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:2c“