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For what values of $x$ is the rate of increase of $x^{3}-2x^{2}+3x+8$ is twice the rate of increase of $x$

\[\begin{array}{1 1}(1)\bigg(-\frac{1}{3},-3\bigg)&(2)\bigg(\frac{1}{3},3\bigg)\\(3)\bigg(-\frac{1}{3},3\bigg)&(4)\bigg(\frac{1}{3},1\bigg)\end{array}\]

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Let $y=x^3-2x^2+3x+8$
Given $\large\frac{dy}{dt} $$=2 \large\frac{dx}{dt}$
$y= x^3-2x^2+3x+8$
$\large\frac{dy}{dx} $$=(3x^2-4x+3) $$\large\frac{dx}{dt}$
$2 \large\frac{dx}{dt} $$=(3x^2-4x+3) $$\large\frac{dx}{dt}$
$2= 3x^2-4x+3$
$3x(x-1) -1(x-1) =0$
$(3x-1) (x-1)=0$
$x=1$ or $x=\large\frac{1}{3}$
Hence 4 is the correct answer
answered May 19, 2014 by meena.p

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