Lösung 1.3:3c

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The only points which can possibly be local extreme points of the function are one of the following:
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The only points which can possibly be local extreme points of the function are one of the following,
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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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What determines the function's region of definition is <math>\ln x</math>, which is defined for <math>x > 0</math>, and this region does not have any endpoints (<math>x=0</math> does not satisfy <math>x>0</math>), so item 3 above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of <math>x</math> and <math>\ln x</math> which are differentiable functions; so, item 2 above does not contribute any extreme points either.
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3. Endpoints of the interval of definition.
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What determines the function's region of definition is
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<math>\ln x</math>, which is defined for
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<math>x>0</math>, and this region does not have any endpoints (
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<math>x=0</math>
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does not satisfy
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<math>x>0</math>
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), so item 3. above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of
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<math>x</math>
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and
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<math>\ln x</math>
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which are differentiable functions; so, item 2. above does not contribute any extreme points either.
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All the remains are possibly critical points. We differentiate the function
All the remains are possibly critical points. We differentiate the function
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{{Displayed math||<math>f^{\,\prime}(x) = 1\cdot \ln x + x\cdot \frac{1}{x} - 0 = \ln x+1</math>}}
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<math>{f}'\left( x \right)=1\centerdot \ln x+x\centerdot \frac{1}{x}-0=\ln x+1</math>
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and see that the derivative is zero when
and see that the derivative is zero when
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{{Displayed math||<math>\ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}</math>}}
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<math>\ln x=-1\quad \Leftrightarrow \quad x=e^{-1}</math>
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In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, <math>f^{\,\prime\prime}(x) = 1/x</math>, which gives that
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In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative,
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<math>{f}''\left( x \right)={1}/{x}\;</math>
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which gives that
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<math>{f}''\left( e^{-1} \right)=\frac{1}{e^{-1}}=e>0</math>
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{{Displayed math||<math>f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,</math>}}
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which implies that
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which implies that <math>x=e^{-1}</math> is a local minimum.
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<math>x=e^{-1}</math>
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is a local minimum.
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Version vom 07:28, 20. Okt. 2008

The only points which can possibly be local extreme points of the function are one of the following,

  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.

What determines the function's region of definition is \displaystyle \ln x, which is defined for \displaystyle x > 0, and this region does not have any endpoints (\displaystyle x=0 does not satisfy \displaystyle x>0), so item 3 above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of \displaystyle x and \displaystyle \ln x which are differentiable functions; so, item 2 above does not contribute any extreme points either.

All the remains are possibly critical points. We differentiate the function

\displaystyle f^{\,\prime}(x) = 1\cdot \ln x + x\cdot \frac{1}{x} - 0 = \ln x+1

and see that the derivative is zero when

\displaystyle \ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}

In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, \displaystyle f^{\,\prime\prime}(x) = 1/x, which gives that

\displaystyle f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,

which implies that \displaystyle x=e^{-1} is a local minimum.

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