Answer 1.3:3

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
Current revision (13:58, 16 September 2008) (edit) (undo)
m (Svar 1.3:3 moved to Answer 1.3:3: Robot: moved page)
 
(One intermediate revision not shown.)
Line 1: Line 1:
{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="50%"| <math>x=0\,</math> (lokal maximipunkt)
+
|width="50%"| <math>x=0\,</math> (local maxima)
|b)
|b)
-
|width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> (lokal minimipunkt)
+
|width="50%"| <math>x=-\frac{1}{3}\ln\frac{5}{3}\,</math> (local minima)
|-
|-
|-
|-
|c)
|c)
-
|width="50%"| <math>x=1/e\,</math> (lokal minimipunkt)
+
|width="50%"| <math>x=1/e\,</math> (local minima)
|d)
|d)
|width="50%"|
|width="50%"|
-
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (lokal maximipunkt)
+
<math>x=-\sqrt{\sqrt{2}-1}\,</math> (local maxima)
-
<math>x=0\,</math> (lokal minimipunkt)
+
<math>x=0\,</math> (local minima)
-
<math>x=\sqrt{\sqrt{2}-1}\,</math> (lokal maximipunkt)
+
<math>x=\sqrt{\sqrt{2}-1}\,</math> (local maxima)
|-
|-
|e)
|e)
-
|width="50%"| <math>x=-3\,</math> (lokal minimipunkt)
+
|width="50%"| <math>x=-3\,</math> (local minima)
-
<math>x=-2\,</math> (lokal maximipunkt)
+
<math>x=-2\,</math> (local maxima)
-
<math>x=1\,</math> (lokal minimipunkt)
+
<math>x=1\,</math> (local minima)
-
<math>x=3\,</math> (lokal maximipunkt)
+
<math>x=3\,</math> (local maxima)
|}
|}

Current revision

a) \displaystyle x=0\, (local maxima) b) \displaystyle x=-\frac{1}{3}\ln\frac{5}{3}\, (local minima)
c) \displaystyle x=1/e\, (local minima) d)

\displaystyle x=-\sqrt{\sqrt{2}-1}\, (local maxima)

\displaystyle x=0\, (local minima)

\displaystyle x=\sqrt{\sqrt{2}-1}\, (local maxima)

e) \displaystyle x=-3\, (local minima)

\displaystyle x=-2\, (local maxima)

\displaystyle x=1\, (local minima)

\displaystyle x=3\, (local maxima)

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Answer_1.3:3"