The correct option is B 4 , 0
Given function y=x3−3x+2
dydx=3x2−3.......(1)
At stationary points, dydx=0,
⇒3x2−3=0
⇒3(x2−1)=0
⇒(x−1)(x+1)=0
⇒x=1,x=−1
Differentiating the equation on both sides again,
d2ydx2=6x .....(2)
Putting x=1 in equation (2)
d2ydx2=6>0
⇒ At point x=1 , local minimum occurs
and local minimum is given by
y(1)=(1)3−3(1)+2=0
Now, putting x=−1 in equation (2)
d2ydx2=−6<0
So, at point x=−1, local maximum occurs and local maximum is given by y(−1)=(−1)3−3(−1)+2=4
So, local maximum and minimum are 4 and 0, respectively