Lösung 1.3:2d

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If the function has several local extreme points, then they must lie between the following three types of points:
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If the function has several local extreme points, then they must be among the following three types of points:
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1. Critical points, i.e. where
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# critical points, i.e. where <math>f^{\,\prime}(x)=0</math>,
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<math>{f}'\left( x \right)=0</math>;
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# points where the function is not differentiable, and
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# endpoints of the interval of definition.
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2. Points where the function is not differentiable;
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Items 2 and 3 do not give rise to any points, because our function, which is a polynomial, is defined and differentiable everywhere. In order to investigate if there are any critical points, we set the derivative,
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3. Endpoints of the interval of definition.
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2. and 3. do not give any points, because our function, which is a polynomial, is defined and differentiable everywhere. In order to investigate if there are any critical points, we set the derivative,
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<math>{f}'\left( x \right)=3x^{2}-18x+30=3\left( x^{2}-6x+10 \right)</math>
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{{Displayed math||<math>f^{\,\prime}(x) = 3x^2 - 18x + 30 = 3(x^2-6x+10)</math>}}
equal to zero and obtain the equation
equal to zero and obtain the equation
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{{Displayed math||<math>x^{2}-6x+10=0\,\textrm{.}</math>}}
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<math>x^{2}-6x+10=0</math>
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Completing the square gives
Completing the square gives
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{{Displayed math||<math>(x-3)^2 - 3^2 + 10 = 0\,,</math>}}
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<math>\left( x-3 \right)^{2}-3^{2}+10=0</math>
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i.e.
i.e.
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{{Displayed math||<math>(x-3)^2 + 1 = 0\,\textrm{.}</math>}}
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<math>\left( x-3 \right)^{2}+1=0</math>
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This equation does not have any real roots, because the left-hand side is always greater than or equal to <math>1</math>, regardless of how <math>x</math> is chosen (the square <math>(x-3)^2</math> can never be negative).
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This equation does not have any real roots, because the left-hand side is always greater than or equal to
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<math>\text{1}</math>, regardless of how
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<math>x</math>
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is chosen (the square
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<math>\left( x-3 \right)^{2}</math>
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can never be negative).
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This means that the function does not have any local extreme points.
This means that the function does not have any local extreme points.
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From the derivative's appearance,
From the derivative's appearance,
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{{Displayed math||<math>f^{\,\prime}(x) = 3((x-3)^2+1)</math>}}
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<math>{f}'\left( x \right)=3\left( \left( x-3 \right)^{2}+1 \right)</math>
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we see that it is always greater than zero and therefore that the function is strictly increasing. We do not have so much more information when we sketch the graph of the function, other than the function's value at a few points.
we see that it is always greater than zero and therefore that the function is strictly increasing. We do not have so much more information when we sketch the graph of the function, other than the function's value at a few points.
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PICTURE GRAPH
 
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[[Image:1_3_2_d.gif|center]]
[[Image:1_3_2_d.gif|center]]

Version vom 13:10, 17. Okt. 2008

If the function has several local extreme points, then they must be among the following three types of points:

  1. critical points, i.e. where \displaystyle f^{\,\prime}(x)=0,
  2. points where the function is not differentiable, and
  3. endpoints of the interval of definition.

Items 2 and 3 do not give rise to any points, because our function, which is a polynomial, is defined and differentiable everywhere. In order to investigate if there are any critical points, we set the derivative,

\displaystyle f^{\,\prime}(x) = 3x^2 - 18x + 30 = 3(x^2-6x+10)

equal to zero and obtain the equation

\displaystyle x^{2}-6x+10=0\,\textrm{.}

Completing the square gives

\displaystyle (x-3)^2 - 3^2 + 10 = 0\,,

i.e.

\displaystyle (x-3)^2 + 1 = 0\,\textrm{.}

This equation does not have any real roots, because the left-hand side is always greater than or equal to \displaystyle 1, regardless of how \displaystyle x is chosen (the square \displaystyle (x-3)^2 can never be negative).

This means that the function does not have any local extreme points.

From the derivative's appearance,

\displaystyle f^{\,\prime}(x) = 3((x-3)^2+1)

we see that it is always greater than zero and therefore that the function is strictly increasing. We do not have so much more information when we sketch the graph of the function, other than the function's value at a few points.

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