Find x-coordinate of stationary point
Find the x-coordinate of the stationary point of the curve $\displaystyle y=\frac{(x-1)^3}{x+1}$, $\displaystyle x>0$
What I did
$\displaystyle \frac{dy}{dx}=\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}$
Stationary Point, dy/dx=0
$\displaystyle \frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}=0$
$\displaystyle (x+1)(3)(x-1)^2-(x-1)^3=0$
$\displaystyle 3(x+1)(x-1)^2=(x-1)^3$
$\displaystyle 3(x+1)=(x-1)$
$\displaystyle 3x+3=x-1$
$\displaystyle 2x+4=0$
$\displaystyle x=-2$
however question states x>0 and ans is 1!