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Lösung 1.3:3b

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.3:3b moved to Solution 1.3:3b: Robot: moved page)
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Because the function is defined and differentiable for all
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<center> [[Image:1_3_3b-1(2).gif]] </center>
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<math>x</math>, the function can only have local extreme points at the critical points, i.e. where the derivative is zero.
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For this function, the derivative is given by
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<center> [[Image:1_3_3b-2(2).gif]] </center>
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<math>{f}'\left( x \right)=-3e^{-3x}+5</math>
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and if we set it to zero, we will obtain
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<math>3e^{-3x}=5</math>
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which is a first-degree equation in
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<math>e^{-3x}</math>
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and has the solution
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<math>x=-\frac{1}{3}\ln \frac{5}{3}</math>
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The function therefore has a critical point
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<math>x=-\frac{1}{3}\ln \frac{5}{3}</math>
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Instead of studying the sign changes of the derivative around the critical point in order to decide if the point is a local maximum, minimum or neither, we investigate the function's second derivative
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The second derivative is equal to
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<math>{f}''\left( x \right)=-3\centerdot \left( -3 \right)e^{-3x}=9e^{-3x}</math>
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and is positive for all values of
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<math>x</math>, since the exponential function is always positive.
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In particular, this means that
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<math>{f}''\left( -\frac{1}{3}\ln \frac{5}{3} \right)>0</math>
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which means that
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<math>x=-\frac{1}{3}\ln \frac{5}{3}</math>
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is a local minimum.

Version vom 11:40, 15. Okt. 2008

Because the function is defined and differentiable for all x, the function can only have local extreme points at the critical points, i.e. where the derivative is zero.

For this function, the derivative is given by


fx=−3e−3x+5 


and if we set it to zero, we will obtain


3e−3x=5


which is a first-degree equation in e−3x and has the solution


x=−31ln35


The function therefore has a critical point x=−31ln35

Instead of studying the sign changes of the derivative around the critical point in order to decide if the point is a local maximum, minimum or neither, we investigate the function's second derivative

The second derivative is equal to


fx=−3−3e−3x=9e−3x 


and is positive for all values of x, since the exponential function is always positive.

In particular, this means that


f−31ln350 


which means that x=−31ln35 is a local minimum.

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