Lösung 1.3:3c
Aus Online Mathematik Brückenkurs 2
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
Zeile 9: | Zeile 9: | ||
All the remains are possibly critical points. We differentiate the function | All the remains are possibly critical points. We differentiate the function | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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 | ||
- | {{ | + | {{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,
- critical points, i.e. where \displaystyle f^{\,\prime}(x) = 0\,,
- points where the function is not differentiable, and
- 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.