Antwort 1.3:3
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
(Der Versionsvergleich bezieht 10 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> ( | + | |width="50%"| <math>x=0\,</math> (Lokales Maximum) |
|b) | |b) | ||
- | |width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> ( | + | |width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> (Lokales Minimum) |
+ | |- | ||
|- | |- | ||
|c) | |c) | ||
- | |width="50%"| <math>x=1/e\,</math> ( | + | |width="50%"| <math>x=1/e\,</math> (Lokales Minimum) |
|d) | |d) | ||
- | |width="50%"| | + | |width="50%"| |
- | <math>x=0\,</math> ( | + | |
- | <math>x=\sqrt{\sqrt{2}-1}\,</math> ( | + | <math>x=-\sqrt{\sqrt{2}-1}\,</math> (Lokales Maximum) |
+ | |||
+ | <math>x=0\,</math> (Lokales Minimum) | ||
+ | |||
+ | <math>x=\sqrt{\sqrt{2}-1}\,</math> (Lokales Maximum) | ||
|- | |- | ||
|e) | |e) | ||
- | |width="50%"| <math>x=-3\,</math> ( | + | |width="50%"| <math>x=-3\,</math> (Lokales Minimum) |
- | <math>x=-2\,</math> ( | + | <math>x=-2\,</math> (Lokales Maximum) |
- | + | ||
- | <math>x=1\,</math> ( | + | <math>x=1\,</math> (Lokales Minimum) |
- | + | ||
- | <math>x=3\,</math> ( | + | <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) |