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

Answered by liberty last updated on 09/Jul/21

$$f\left(x\right)=x^3+{ax}^2+bx+c;a,b,c\in R$$$$f{}^{\prime }\left(x\right)=3x^2+2ax+b$$$$f{}^{″}\left(x\right)=6x+2a$$$$\Rightarrow g\left(x\right)=x^3+(a+3)x^2+(6+2a+b)x+2a+b+c$$$$whereg{}^{\prime }\left(x\right)=0forx=-3\wedge x=6$$$$g^{\prime }\left(x\right)=3x^2+2(a+3)x+(6+2a+b)=0$$$$\Rightarrow x_1+x_2=3=-\frac{2(a+3)}{3}$$$$\Rightarrow -\frac{9}{2}-3=a;a=-\frac{15}{2}$$$$\Rightarrow x_1.x_2=-18=\frac{6+2a+b}{3}$$$$\Rightarrow -54=6-15+b,b=-45$$$$g\left(x\right)=x^3-\frac{9}{2}{x}^2-54x-60+c$$$$...$$$$\Rightarrow y=1\wedge y=\frac{f\left(x\right)}{g\left(x\right)+6}$$$$\Rightarrow f\left(x\right)=g\left(x\right)+6$$$$\Rightarrow x^3-\frac{15}{2}{x}^2-45x+c=x^3-\frac{9}{2}{x}^2-54x+c-54$$$$\Rightarrow 3x^2-9x-54=0$$$$\Rightarrow x^2-3x-18=0$$$$\Rightarrow (x-6)(x+3)=0;x=-3\wedge x=6$$$$area=\underset{-3}{\stackrel{6}{\int }}(\frac{f\left(x\right)}{g\left(x\right)+6}-1)dx$$$$$$

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