From Förberedande kurs i matematik 2

Revision as of 07:54, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (12:46, 17 October 2008) (edit) (undo) Tek (Talk | contribs) m
(2 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	In order to determine the function's extreme points, we investigate three types of points,
-	<center> [[Image:1_3_2b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:1_3_2b-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	[[Image:1_3_2_b.gif|center]]	+	# critical points, i.e. where <math>f^{\,\prime}(x)=0</math>,
		+	# points where the function is not differentiable, and
		+	# endpoints of the interval of definition.
		+
		+	In our case, we have that:
		+
		+	<ol>
		+	<li>The derivative of <math>f(x)</math> is given by
		+	{{Displayed math||<math>f^{\,\prime}(x) = 3-2x</math>}}
		+	and becomes zero when <math>x=3/2\,</math>.</li>
		+
		+	<li>The function is a polynomial, and is therefore differentiable everywhere.</li>
		+
		+	<li>The function is defined for all ''x'', and therefore the interval of definition has no endpoints.</li>
		+	</ol>
		+
		+	There is thus just one point <math>x=3/2\,</math>, where the function possibly has an extreme point.
		+
		+	If we write down a sign table for the derivative, we see that <math>x=3/2</math> is a local maximum.
		+
		+
		+	{| border="1" cellpadding="5" cellspacing="0" align="center"
		+	|-
		+	|width="50px" align="center" style="background:#efefef;"| <math>x</math>
		+	|width="50px" align="center" style="background:#efefef;"|
		+	|width="50px" align="center" style="background:#efefef;"| <math>\tfrac{3}{2}</math>
		+	|width="50px" align="center" style="background:#efefef;"|
		+	|-
		+	|width="50px" align="center"| <math>f^{\,\prime}(x)</math>
		+	|width="50px" align="center"| <math>+</math>
		+	|width="50px" align="center"| <math>0</math>
		+	|width="50px" align="center"| <math>-</math>
		+	|-
		+	|width="50px" align="center"| <math>f(x)</math>
		+	|width="50px" align="center"| <math>\nearrow</math>
		+	|width="50px" align="center"| <math>\tfrac{17}{4}</math>
		+	|width="50px" align="center"| <math>\searrow</math>
		+	|}
		+
		+
		+	Because the function is given by a second-degree expression, its graph is a parabola with a maximum at <math>(3/2, 17/4)</math> and we can draw it with the help of a few couple of points.
		+
		+	[[Image:1_3_2_b.gif||center]]

---

Current revision

In order to determine the function's extreme points, we investigate three types of points,

1. critical points, i.e. where \displaystyle f^{\,\prime}(x)=0,
2. points where the function is not differentiable, and
3. endpoints of the interval of definition.

In our case, we have that:

1. The derivative of \displaystyle f(x) is given by

   \displaystyle f^{\,\prime}(x) = 3-2x

   and becomes zero when \displaystyle x=3/2\,.
2. The function is a polynomial, and is therefore differentiable everywhere.
3. The function is defined for all x, and therefore the interval of definition has no endpoints.

There is thus just one point \displaystyle x=3/2\,, where the function possibly has an extreme point.

If we write down a sign table for the derivative, we see that \displaystyle x=3/2 is a local maximum.

\displaystyle x		\displaystyle \tfrac{3}{2}
\displaystyle f^{\,\prime}(x)	\displaystyle +	\displaystyle 0	\displaystyle -
\displaystyle f(x)	\displaystyle \nearrow	\displaystyle \tfrac{17}{4}	\displaystyle \searrow

Because the function is given by a second-degree expression, its graph is a parabola with a maximum at \displaystyle (3/2, 17/4) and we can draw it with the help of a few couple of points.