Antwort 1.3:3

Aus Online Mathematik Brückenkurs 2

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(Translated into English)
Aktuelle Version (09:54, 5. Aug. 2009) (bearbeiten) (rückgängig)
 
(Der Versionsvergleich bezieht 3 dazwischen liegende Versionen mit ein.)
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{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="50%"| <math>x=0\,</math> (local maxima)
+
|width="50%"| <math>x=0\,</math> (Lokales Maximum)
|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 Minimum)
|-
|-
|-
|-
|c)
|c)
-
|width="50%"| <math>x=1/e\,</math> (local minima)
+
|width="50%"| <math>x=1/e\,</math> (Lokales Minimum)
|d)
|d)
|width="50%"|
|width="50%"|
-
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (local maxima)
+
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (Lokales Maximum)
-
<math>x=0\,</math> (local minima)
+
<math>x=0\,</math> (Lokales Minimum)
-
<math>x=\sqrt{\sqrt{2}-1}\,</math> (local maxima)
+
<math>x=\sqrt{\sqrt{2}-1}\,</math> (Lokales Maximum)
|-
|-
|e)
|e)
-
|width="50%"| <math>x=-3\,</math> (local minima)
+
|width="50%"| <math>x=-3\,</math> (Lokales Minimum)
-
<math>x=-2\,</math> (local maxima)
+
<math>x=-2\,</math> (Lokales Maximum)
-
<math>x=1\,</math> (local minima)
+
<math>x=1\,</math> (Lokales Minimum)
-
<math>x=3\,</math> (local maxima)
+
<math>x=3\,</math> (Lokales Maximum)
|}
|}

Aktuelle Version

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

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

\displaystyle x=0\, (Lokales Minimum)

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

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

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

\displaystyle x=1\, (Lokales Minimum)

\displaystyle x=3\, (Lokales Maximum)

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