Lösung 1.3:3c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.3:3c moved to Solution 1.3:3c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
The only points which can possibly be local extreme points of the function are one of the following:
-
<center> [[Image:1_3_3c-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
1. Critical points, i.e. where
-
{{NAVCONTENT_START}}
+
<math>{f}'\left( x \right)=0</math>;
-
<center> [[Image:1_3_3c-2(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
2. Points where the function is not differentiable;
 +
 
 +
3. Endpoints of the interval of definition.
 +
 
 +
What determines the function's region of definition is
 +
<math>\ln x</math>, which is defined for
 +
<math>x>0</math>, and this region does not have any endpoints (
 +
<math>x=0</math>
 +
does not satisfy
 +
<math>x>0</math>
 +
), 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
 +
<math>x</math>
 +
and
 +
<math>\ln x</math>
 +
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
 +
 
 +
 
 +
<math>{f}'\left( x \right)=1\centerdot \ln x+x\centerdot \frac{1}{x}-0=\ln x+1</math>
 +
 
 +
 
 +
and see that the derivative is zero when
 +
 
 +
 
 +
<math>\ln x=-1\quad \Leftrightarrow \quad x=e^{-1}</math>
 +
 
 +
 
 +
 
 +
In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative,
 +
<math>{f}''\left( x \right)={1}/{x}\;</math>
 +
which gives that
 +
 
 +
 
 +
<math>{f}''\left( e^{-1} \right)=\frac{1}{e^{-1}}=e>0</math>
 +
 
 +
 
 +
which implies that
 +
<math>x=e^{-1}</math>
 +
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 \displaystyle {f}'\left( x \right)=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 \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}'\left( x \right)=1\centerdot \ln x+x\centerdot \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}


In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, \displaystyle {f}''\left( x \right)={1}/{x}\; which gives that


\displaystyle {f}''\left( e^{-1} \right)=\frac{1}{e^{-1}}=e>0


which implies that \displaystyle x=e^{-1} is a local minimum.

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