Lösung 1.3:2b

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In order to determine the function's extreme points, we investigate three types of points:
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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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2. Points where the function is not differentiable;
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3. Endpoints of the interval of definition.
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In our case, we have that:
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1. The derivative of
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<math>f\left( x \right)</math>
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is given by
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<math>{f}'\left( x \right)=3-2x</math>
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and becomes zero when
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<math>x=\frac{3}{2}</math>.
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2. The function is a polynomial, and is therefore differentiable everywhere.
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3. The function is defined for all
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<math>x</math>, and there are therefore the interval of definition has no
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endpoints.
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There is thus a point
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<math>x=\frac{3}{2}</math>, where the function possibly has an extreme point.
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If we write down a sign table for the derivative, we see that
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<math>x=\frac{3}{2}</math>
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is a local maximum.
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TABLE
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Because the function is given by a second-degree expression, its graph is a parabola with a maximum at
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<math>\left( \frac{3}{2} \right.,\left. \frac{17}{4} \right)</math>
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and we can draw it with the help of a few couple of points.
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PICTURE TABLE

Version vom 10:18, 15. Okt. 2008

In order to determine the function's extreme points, we investigate three types of 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.

In our case, we have that:

1. The derivative of \displaystyle f\left( x \right) is given by

\displaystyle {f}'\left( x \right)=3-2x

and becomes zero when \displaystyle x=\frac{3}{2}.

2. The function is a polynomial, and is therefore differentiable everywhere.

3. The function is defined for all \displaystyle x, and there are therefore the interval of definition has no endpoints.

There is thus a point \displaystyle x=\frac{3}{2}, where the function possibly has an extreme point.

If we write down a sign table for the derivative, we see that \displaystyle x=\frac{3}{2} is a local maximum.

TABLE

Because the function is given by a second-degree expression, its graph is a parabola with a maximum at \displaystyle \left( \frac{3}{2} \right.,\left. \frac{17}{4} \right) and we can draw it with the help of a few couple of points.

PICTURE TABLE

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