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Lösung 1.3:3c

Aus Online Mathematik Brückenkurs 2

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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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<center> [[Image:1_3_3c-1(2).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_3c-2(2).gif]] </center>
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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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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
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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
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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,
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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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which implies that
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<math>x=e^{-1}</math>
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is a local minimum.

Version vom 11:54, 15. 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 fx=0 ;

2. Points where the function is not differentiable;

3. Endpoints of the interval of definition.

What determines the function's region of definition is lnx, which is defined for x0, and this region does not have any endpoints ( x=0 does not satisfy x0 ), 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 x and lnx 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


fx=1lnx+xx1−0=lnx+1 


and see that the derivative is zero when


lnx=−1x=e−1


In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, fx=1x  which gives that


fe−1=1e−1=e0 


which implies that x=e−1 is a local minimum.

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