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		<title>3.1 Lösning 9a - Versionshistorik</title>
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		<updated>2026-11-05T05:31:13Z</updated>
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		<title>Taifun den 5 december 2014 kl. 23.59</title>
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				<updated>2014-12-05T23:59:19Z</updated>
		
		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;Vi har:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; f(x) = \, x^3 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; f\,'(x) = 3\,x^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Medelvärdessatsen:&lt;br /&gt;
&lt;br /&gt;
:Det finns minst en punkt &amp;lt;math&amp;gt; \, c \, &amp;lt;/math&amp;gt; i intervallet &amp;lt;math&amp;gt; \, 1 &amp;lt; x &amp;lt; 3 \, &amp;lt;/math&amp;gt; så att det gäller:&lt;br /&gt;
&lt;br /&gt;
::::&amp;lt;math&amp;gt; \begin{array}{rcl}   {f(3) \, - \, f(1) \over 3 - 1} &amp;amp; = &amp;amp; f\,'(c)   \\&lt;br /&gt;
                                                                                \\&lt;br /&gt;
                                  {3^3 \, - \, 1^3 \over 3 - 1} &amp;amp; = &amp;amp; 3\,c^2    \\&lt;br /&gt;
                                                                                \\&lt;br /&gt;
                                         {27 \, - \, 1 \over 2} &amp;amp; = &amp;amp; 3\,c^2    \\&lt;br /&gt;
                                                                                \\&lt;br /&gt;
                                                             13 &amp;amp; = &amp;amp; 3\,c^2   &lt;br /&gt;
           \end{array} &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Derivatans medelvärde i intervallet &amp;lt;math&amp;gt; \, 1 \leq x \leq 3 \, &amp;lt;/math&amp;gt; är &amp;lt;math&amp;gt; \, 13 \, &amp;lt;/math&amp;gt;.&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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