Antwort 1.3:3

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{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
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|width="50%"| <math>x=0\,</math> (local maxima)
+
|width="50%"| <math>x=0\,</math> (Lokales Maxima)
|b)
|b)
-
|width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> (local minima)
+
|width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> (Lokales Minima)
|-
|-
|-
|-
|c)
|c)
-
|width="50%"| <math>x=1/e\,</math> (local minima)
+
|width="50%"| <math>x=1/e\,</math> (Lokales Minima)
|d)
|d)
|width="50%"|
|width="50%"|
-
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (local maxima)
+
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (Lokales Maxima)
-
<math>x=0\,</math> (local minima)
+
<math>x=0\,</math> (Lokales Minima)
-
<math>x=\sqrt{\sqrt{2}-1}\,</math> (local maxima)
+
<math>x=\sqrt{\sqrt{2}-1}\,</math> (Lokales Maxima)
|-
|-
|e)
|e)
-
|width="50%"| <math>x=-3\,</math> (local minima)
+
|width="50%"| <math>x=-3\,</math> (Lokales Minima)
-
<math>x=-2\,</math> (local maxima)
+
<math>x=-2\,</math> (Lokales Maxima)
-
<math>x=1\,</math> (local minima)
+
<math>x=1\,</math> (Lokales Minima)
-
<math>x=3\,</math> (local maxima)
+
<math>x=3\,</math> (Lokales Maxima)
|}
|}

Version vom 14:38, 26. Apr. 2009

a) \displaystyle x=0\, (Lokales Maxima) b) \displaystyle x=-\frac{1}{3}\ln\frac{5}{3}\, (Lokales Minima)
c) \displaystyle x=1/e\, (Lokales Minima) d)

\displaystyle x=-\sqrt{\sqrt{2}-1}\, (Lokales Maxima)

\displaystyle x=0\, (Lokales Minima)

\displaystyle x=\sqrt{\sqrt{2}-1}\, (Lokales Maxima)

e) \displaystyle x=-3\, (Lokales Minima)

\displaystyle x=-2\, (Lokales Maxima)

\displaystyle x=1\, (Lokales Minima)

\displaystyle x=3\, (Lokales Maxima)

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/Antwort_1.3:3“