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		<title>3.1 Lösning 9d - Versionshistorik</title>
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		<updated>2026-11-05T06:52:25Z</updated>
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	<entry>
		<id>https://wiki.mattekollen.se/index.php?title=3.1_L%C3%B6sning_9d&amp;diff=18663&amp;oldid=prev</id>
		<title>Taifun den 5 december 2014 kl. 14.52</title>
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				<updated>2014-12-05T14:52:25Z</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;
:Punkterna &amp;lt;math&amp;gt; \, (1, f(1)) \, &amp;lt;/math&amp;gt; och &amp;lt;math&amp;gt; \, (3, f(3)) \, &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sekanten&amp;lt;span style=&amp;quot;color:black&amp;quot;&amp;gt;:&amp;lt;/span&amp;gt; &amp;lt;math&amp;gt; {\color{White} x} y \, = \, k\,x \, + \, m &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sekantens lutning:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; k = {f(3) - f(1) \over 3 - 1} = {3^3 - 1^3 \over 2} = {27 - 1 \over 2} = {26 \over 2} = 13 &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sekantens ekvation&amp;lt;span style=&amp;quot;color:black&amp;quot;&amp;gt;:&amp;lt;/span&amp;gt; &amp;lt;math&amp;gt; y \, = \, 13\,x \, + \, m &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Punkten &amp;lt;math&amp;gt; \, (1, f(1)) \, = \, (1, 1) \, &amp;lt;/math&amp;gt; ligger på sekanten:&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;\begin{array}{rcl}  y &amp;amp; = &amp;amp; 13\,x \, + \, m       \\&lt;br /&gt;
                           1 &amp;amp; = &amp;amp; 13 \cdot 1 \, + \, m  \\&lt;br /&gt;
                           1 &amp;amp; = &amp;amp; 13 \, + \, m          \\&lt;br /&gt;
                         -12 &amp;amp; = &amp;amp; m                     \\&lt;br /&gt;
       \end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sekantens ekvation:&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;math&amp;gt; y \, = \, 13\,x \, - \, 12 &amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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