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Por defecto en DIFERENCIAL C (161)
EXAMEN 1 (15)
THOMAS 136-6 variante 2 (300)
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aa: rand_with_prohib(-15,15,[-1,0,1]); bb: rand_with_prohib(-15,15,[-1,0,1]); cc: rand_with_prohib(-15,15,[-1,0,1]); dd: rand_with_prohib(-15,15,[-1,0,1]); exp:aa*x^3+bb*x^2+cc*x+dd; pt:1; ta1:diff(exp,x); ta2:subst(x=pt,ta1); ta3:remainder(exp,(x-pt)^2); ta4:subst(x=pt,ta2); ta5:subst(x=pt,exp);
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<p></p><p></p><p>\[\large{\textrm{En este ejercicio, calcular\'a la ecuaci\'on de la recta tangente}}\]</p><p>\[\large{\mathlarger{y(x)}\ \textrm{de la funci\'on:} \\ \ \mathlarger{f(x)={@exp@}}\ \textrm{en el punto}\ \textstyle{\Large{(1,{@ta5@})}}.}\]. </p><p><br>\[\large{\textrm{Siguiendo los pasos:}}\]</p><p><br></p><p></p><p> \[\large{\textrm{1.- Calcule la derivada}\ \ \ f^{\ \prime}\left(x\right) \ = \ \ }\] [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p><p>\[\large{\textrm{2.- Calcule }\ \ \ f^{\ \prime}\left(1\right) \ = \ \ }\] [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p><p> \[\large{\textrm{3.- Calcule la ecuaci\'on de la recta tangente}\ \ \ y\left(x\right) \ = \ \ }\] [[input:ans3]] [[validation:ans3]] [[feedback:prt3]]</p><p><br></p>
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<p>Seguiremos los pasos indicados.</p><p>1.- Calculemos la derivada de \(\displaystyle f(x)={@exp@}\).</p><p><br></p><p>\(\displaystyle\frac{d}{dx}\left({@exp@}\right)=\)</p><p>\(\displaystyle = \frac{d}{dx}\left({@coeff(exp,x,3)@}x^3\right)+\frac{d}{dx}\left({@coeff(exp,x,2)@}x^2\right)+\frac{d}{dx}\left({@coeff(exp,x,1)@}x\right)+\frac{d}{dx}\left({@coeff(exp,x,0)@}\right)=\) </p><p>\(\displaystyle = ({@coeff(exp,x,3)@})\frac{d}{dx}\left(x^3\right)+({@coeff(exp,x,2)@})\frac{d}{dx}\left(x^2\right)+({@coeff(exp,x,1)@})\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left({@coeff(exp,x,0)@}\right)=\)</p><p>\(\displaystyle =({@coeff(exp,x,3)@})(3)x^2+({@coeff(exp,x,2)@})(2)x+\left({@coeff(exp,x,1)@}\right)=\) </p><p>\(\displaystyle = {@ta1@}\). </p><p><br></p><p>Por tanto \(\displaystyle\frac{d}{dx}\left({@exp@}\right)={@ta1@}\)</p><p><br></p><p></p><p>2.- Calculemos la pendiente de la recta tangente a la función \(\displaystyle f(x)={@exp@}\) en el punto \(\displaystyle (1,{@ta5@})\) .</p><p><br></p><p></p><p>Para ello, primeramente calculamos \(f^{\ \prime}\left(1\right)\ : \)</p><p>Como \(f^{\ \prime}\left(x\right)={@ta1@}\), se sigue que \(f^{\ \prime}\left(1\right)=({@coeff(ta1,x,2)@})(1)^2+({@coeff(ta1,x,1)@})(1)+({@coeff(ta1,x,0)@})=({@coeff(ta1,x,2)@})+({@coeff(ta1,x,1)@})+({@coeff(ta1,x,0)@})={@ta4@}\). </p><p><br></p><p>Por tanto \(f^{\ \prime}\left(1\right)={@ta4@}\).</p><p><br> </p><p></p><p>3.- Calculemos la ecuación de la recta tangente a la función \(\displaystyle f(x)={@exp@}\) en el punto \(\displaystyle (1,{@ta5@})\).<br></p><p>Sabemos que la ecuación de la recta tangente a una función \(g(x)\) en el punto \((a,g(a))\) viene dada por:</p><p>\[y-g(a)= g^{\ \prime}\left(a\right)(x-a)\].</p><p>Por tanto la ecuación de la recta tangente a la función \(\displaystyle f(x)={@exp@}\) en el punto \(\displaystyle (1,{@ta5@})\) es:</p><p> \(y-f(1)= f^{\ \prime}\left(1\right)(x-1)\). </p><p><br></p><p>Calculando:</p><p> \(f(1)=({@coeff(exp,x,3)@})(1)^3+({@coeff(exp,x,2)@})(1)^2+({@coeff(exp,x,1)@})+({@coeff(exp,x,0)@})=({@coeff(exp,x,3)@})+({@coeff(exp,x,2)@})+({@coeff(exp,x,1)@})+({@coeff(exp,x,0)@})={@ta5@}\). </p><p><br></p><p>Como \(f^{\ \prime}\left(1\right)={@ta4@}\) se tiene entones que:</p><p> \(y-f(1)= f^{\ \prime}\left(1\right)(x-1)\), esto es \(y-({@ta5@})= {@ta4@}(x-1)={@ta4@}x-({@ta4@})\).</p><p><br></p><p>Se sigue que la ecuación de la recta tangente es \(y= {@ta4@}x+({@-ta4+ta5@})\)<br></p><p><br></p><p><br></p>
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Dr. Ricardo Lopez
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