Question-145888

Question Number 145888 by bramlexs22 last updated on 09/Jul/21

Answered by liberty last updated on 09/Jul/21

f(x)=x^3 +ax^2 +bx+c ; a,b,c ∈R f ′(x)=3x^2 +2ax+b f ′′(x)=6x+2a ⇒g(x)=x^3 +(a+3)x^2 +(6+2a+b)x+2a+b+c where g ′(x)=0 for x=−3 ∧x=6 g′(x)=3x^2 +2(a+3)x+(6+2a+b)=0 ⇒x_1 +x_2 =3 =−((2(a+3))/3) ⇒−(9/2)−3=a ; a=−((15)/2) ⇒x_1 .x_2 =−18=((6+2a+b)/3) ⇒−54= 6−15+b ,b=−45 g(x)=x^3 −(9/2)x^2 −54x−60+c ... ⇒y=1 ∧ y= ((f(x))/(g(x)+6)) ⇒f(x)=g(x)+6 ⇒x^3 −((15)/2)x^2 −45x+c=x^3 −(9/2)x^2 −54x+c−54 ⇒3x^2 −9x−54=0 ⇒x^2 −3x−18=0 ⇒(x−6)(x+3)=0 ; x=−3 ∧ x=6 area =∫_(−3) ^6 (((f(x))/(g(x)+6))−1)dx

$${f}\left({x}\right)={x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\:;\:{a},{b},{c}\:\in{R} \\ $$$${f}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{ax}+{b} \\ $$$${f}\:''\left({x}\right)=\mathrm{6}{x}+\mathrm{2}{a}\: \\ $$$$\Rightarrow{g}\left({x}\right)={x}^{\mathrm{3}} +\left({a}+\mathrm{3}\right){x}^{\mathrm{2}} +\left(\mathrm{6}+\mathrm{2}{a}+{b}\right){x}+\mathrm{2}{a}+{b}+{c} \\ $$$${where}\:{g}\:'\left({x}\right)=\mathrm{0}\:{for}\:{x}=−\mathrm{3}\:\wedge{x}=\mathrm{6} \\ $$$${g}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\left({a}+\mathrm{3}\right){x}+\left(\mathrm{6}+\mathrm{2}{a}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{3}\:=−\frac{\mathrm{2}\left({a}+\mathrm{3}\right)}{\mathrm{3}} \\ $$$$\Rightarrow−\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}={a}\:;\:{a}=−\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} .{x}_{\mathrm{2}} =−\mathrm{18}=\frac{\mathrm{6}+\mathrm{2}{a}+{b}}{\mathrm{3}} \\ $$$$\Rightarrow−\mathrm{54}=\:\mathrm{6}−\mathrm{15}+{b}\:,{b}=−\mathrm{45}\: \\ $$$${g}\left({x}\right)={x}^{\mathrm{3}} −\frac{\mathrm{9}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{54}{x}−\mathrm{60}+{c}\: \\ $$$$… \\ $$$$\Rightarrow{y}=\mathrm{1}\:\wedge\:{y}=\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)+\mathrm{6}} \\ $$$$\Rightarrow{f}\left({x}\right)={g}\left({x}\right)+\mathrm{6}\: \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\frac{\mathrm{15}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{45}{x}+{c}={x}^{\mathrm{3}} −\frac{\mathrm{9}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{54}{x}+{c}−\mathrm{54} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} −\mathrm{9}{x}−\mathrm{54}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{18}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{6}\right)\left({x}+\mathrm{3}\right)=\mathrm{0}\:;\:{x}=−\mathrm{3}\:\wedge\:{x}=\mathrm{6} \\ $$$${area}\:=\underset{−\mathrm{3}} {\overset{\mathrm{6}} {\int}}\:\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)+\mathrm{6}}−\mathrm{1}\right){dx} \\ $$$$ \\ $$

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