Lösung 1.3:3b

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Because the function is defined and differentiable for all
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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.
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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
For this function, the derivative is given by
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{{Displayed math||<math>f^{\,\prime}(x) = -3e^{-3x} + 5</math>}}
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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
and if we set it to zero, we will obtain
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{{Displayed math||<math>3e^{-3x} = 5</math>}}
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<math>3e^{-3x}=5</math>
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which is a first-degree equation in <math>e^{-3x}</math> and has the solution
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{{Displayed math||<math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>}}
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which is a first-degree equation in
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The function therefore has a critical point <math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>
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<math>e^{-3x}</math>
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and has the solution
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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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<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
The second derivative is equal to
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{{Displayed math||<math>f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x}</math>}}
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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 ''x'', since the exponential function is always positive.
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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
In particular, this means that
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{{Displayed math||<math>f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,,</math>}}
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<math>{f}''\left( -\frac{1}{3}\ln \frac{5}{3} \right)>0</math>
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which means that <math>x=-\tfrac{1}{3}\ln\tfrac{5}{3}</math> is a local minimum.
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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.
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Version vom 14:37, 17. 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

\displaystyle f^{\,\prime}(x) = -3e^{-3x} + 5

and if we set it to zero, we will obtain

\displaystyle 3e^{-3x} = 5

which is a first-degree equation in \displaystyle e^{-3x} and has the solution

\displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}

The function therefore has a critical point \displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}

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

\displaystyle f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x}

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

In particular, this means that

\displaystyle f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,,

which means that \displaystyle x=-\tfrac{1}{3}\ln\tfrac{5}{3} is a local minimum.

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