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		<title>3.1 Lösning 4a - Versionshistorik</title>
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	<entry>
		<id>https://wiki.mattekollen.se/index.php?title=3.1_L%C3%B6sning_4a&amp;diff=18347&amp;oldid=prev</id>
		<title>Taifun den 1 december 2014 kl. 08.44</title>
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				<updated>2014-12-01T08:44:46Z</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;:&amp;lt;math&amp;gt; f(x) \,=\, -3\,x^3 + 27\,x^2 - 45\,x &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt; f'(x) \,=\, -9\,x^2 + 54\,x - 45 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
För att hitta derivatans nollställen sätter vi &amp;lt;math&amp;gt; \, f'(x) \, &amp;lt;/math&amp;gt; till &amp;lt;math&amp;gt; \, 0 \, &amp;lt;/math&amp;gt; och beräknar &amp;lt;math&amp;gt; \, x \, &amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;\begin{array}{rcl} -9\,x^2 + 54\,x - 45 &amp;amp;     =    &amp;amp; 0  \\&lt;br /&gt;
                              x^2 -  6\,x +  5 &amp;amp;     =    &amp;amp; 0  &lt;br /&gt;
        \end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
[[1.2_Repetition_Faktorisering_%26_Vieta_från_Matte_2#Vietas_formler_-_samband_mellan_koefficienter_och_nollst.C3.A4llen|&amp;lt;strong&amp;gt;&amp;lt;span style=&amp;quot;color:blue&amp;quot;&amp;gt;Vieta&amp;lt;/span&amp;gt;&amp;lt;/strong&amp;gt;]]:&lt;br /&gt;
::::&amp;lt;math&amp;gt; \begin{array}{rcl} x_1 \cdot x_2 &amp;amp;     =    &amp;amp; 5         \\&lt;br /&gt;
                              x_1   +   x_2 &amp;amp;     =    &amp;amp; -(-6) = 6 \\&lt;br /&gt;
                                            &amp;amp;\Downarrow&amp;amp;    \\&lt;br /&gt;
                                        x_1 &amp;amp;     =    &amp;amp; 1  \\&lt;br /&gt;
                                        x_2 &amp;amp;     =    &amp;amp; 5&lt;br /&gt;
           \end{array}&amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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