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	<title>Comentários sobre: Cálculo de limite envolvendo função do segundo grau com delta &lt; 0</title>
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		<title>Por: Cassioeduardo</title>
		<link>https://www.professoronline.net/calculo-de-limite-envolvendo-funcao-do-segundo-grau-com-delta-0-3/comment-page-1/#comment-1589</link>
		<dc:creator><![CDATA[Cassioeduardo]]></dc:creator>
		<pubDate>Wed, 28 May 2014 00:32:25 +0000</pubDate>
		<guid isPermaLink="false">#comment-1589</guid>
		<description><![CDATA[Eu fiz as contas e cheguei a esse resultado
f&#039;(x) =[(t² - 1)&#039;(t² + t - 2)-(t² - 1)(t² + t - 2)&#039;]/(t² + t - 2)² 
f&#039;(x) =[2t(t² + t - 2)-(t² - 1)(2t + 1)]/(t² + t - 2)² 
f&#039;(x) =[2t³+ 2t² - 4t)-(2t³ -2t +t²- 1)]/(t² + t - 2)² 
f&#039;(x) =[2t³+ 2t² - 4t-2t³ +2t -t²+1)]/(t² + t - 2)² 
f&#039;(x) =[t² -2t+1]/(t² + t - 2)² 
f&#039;(x) =[t-1]²/(t^4+2t³-3t² -4 t +4) 
f&#039;(x) =[t-1]²/(t³+3t² -4 )(t -1) 
f&#039;(x) =[t-1]/(t³+3t² -4 )]]></description>
		<content:encoded><![CDATA[<p>Eu fiz as contas e cheguei a esse resultado<br />
f'(x) =[(t² &#8211; 1)'(t² + t &#8211; 2)-(t² &#8211; 1)(t² + t &#8211; 2)&#8217;]/(t² + t &#8211; 2)²<br />
f'(x) =[2t(t² + t &#8211; 2)-(t² &#8211; 1)(2t + 1)]/(t² + t &#8211; 2)²<br />
f'(x) =[2t³+ 2t² &#8211; 4t)-(2t³ -2t +t²- 1)]/(t² + t &#8211; 2)²<br />
f'(x) =[2t³+ 2t² &#8211; 4t-2t³ +2t -t²+1)]/(t² + t &#8211; 2)²<br />
f'(x) =[t² -2t+1]/(t² + t &#8211; 2)²<br />
f'(x) =[t-1]²/(t^4+2t³-3t² -4 t +4)<br />
f'(x) =[t-1]²/(t³+3t² -4 )(t -1)<br />
f'(x) =[t-1]/(t³+3t² -4 )</p>
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