Antwort 1.3:3
Aus Online Mathematik Brückenkurs 2
| 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) |
