Question

What are the local extremes of y=2x^4-4x^2+1? Support answer with a graph?

Answer 1

you need to differeintiate the function to get dy/dx=8x^3-8x.
Then you set this equal to zero to find where the critical points are.
So X(8x^2-8)=0
So either x=0 or x^2=1.
If x=0, y=1. so (0,1) is a local max
If x=-1, y=-1 so (-1,-1) is a local min
If X=1, y=-1 so (1, -1) is a local min