Question

Find the coordinates of the points on the curve $y=x^4-2x^2+2$ ,for which first derivative is $0$.

• A. $(0,2),(1,1)$ and $(-1,1)$
• B. $(2,2),(1,1)$ and $(-1,1)$
• C. $(2,0),(1,1)$ and $(-1,1)$
• D. $(0,2),(1,1)$ and $(1,1)$

Answer 1

The correct option is A $(0,2),(1,1)$ and $(-1,1)$

$\frac{dy}{dx}=\frac{d}{dx}(x^4-2x^2+2)\frac{dy}{dx}=4x^3-4x\frac{dy}{dx}=04x^3-4x=04x(x^2-1)=0\Rightarrow x=0,1,-1$
The corresponding values of $y$ are: $2,1,1$

$\therefore$The corresponding points on the curve are $(0,2),(1,1)$ and $(-1,1)$.