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Lösung 1.3:3c

Aus Online Mathematik Brückenkurs 2

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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>}}
+
{{Abgesetzte Formel||<math>f^{\,\prime}(x) = 1\cdot \ln x + x\cdot \frac{1}{x} - 0 = \ln x+1</math>}}
and see that the derivative is zero when
and see that the derivative is zero when
-
{{Displayed math||<math>\ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}</math>}}
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
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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{{Displayed math||<math>f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,</math>}}
+
{{Abgesetzte Formel||<math>f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,</math>}}
which implies that <math>x=e^{-1}</math> is a local minimum.
which implies that <math>x=e^{-1}</math> is a local minimum.

Version vom 12:56, 10. Mär. 2009

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

  1. critical points, i.e. where f(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 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

f(x)=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, f(x)=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“