Question

if the rate of increase of x cube -5x square +5x +8 is twice the rate of increase of x , then the values of x are?​

Answer 1

Answer:

3 and 1/3

Step-by-step explanation:

Let

$f(x)=x^3-5x^2+5x+8$

Rate of increase of f(x) is given by:

$f'(x)= 3x^2-10x+5$

Let

$g(x)=x$

Rate of increase of g(x) is given by:

$g'(x)= 1$

As the rate of increase of f(x) is twice the rate of increase of g(x),

$f'(x)=2(g'(x))$

$3x^2-10x+5=2(1)$

$3x^2-10x+5-2=0\\3x^2-10x+3=0\\3x^2-9x-x+3=0\\3x(x-3)-1(x-3)=0\\(3x-1)(x-3)=0$

Therefore,

$x=3,\frac{1}{3}$