Question

$y=x^3-x^2-x+7$ find x =? when $\frac{dy}{dx}=0$

Answer 1

Given

$y=x^3-x^2-x+7$

Differentiating it w.r.t x

$\frac{dy}{dx}=3x^2-2x-1$

$\frac{dy}{dx}=0$ [Given]

$3x^2-2x-1=0$

$3x^2-3x+x-1=0$

$3x(x-1)+1(x-1)=0$

$(3x+1)(x-1)=0$

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