How do you find the local max and min for f(x) = x^4 - 2x^2 + 1f(x)=x42x2+1?

1 Answer
Sep 26, 2016

max at (0 ,1), mins at (-1 ,0) and (1 ,0)

Explanation:

Differentiate f(x) and equate to zero to find color(blue)"critical points"critical points

rArrf'(x)=4x^3-4x

4x^3-4x=0rArr4x(x^2-1)=0rArr4x(x-1)(x+1)=0

rArrx=0,x=-1,x=1

Find the y coordinates.

f(0)=0-0+1=1rArr(0,1)

f(-1)=1-2+1=0rArr(-1,0)

f(1)=1-2+1=0rArr(1,0)

To find if max/min use the color(blue)"second derivative test"

• If f''(a) > 0 , then minimum

• If f''(a) < 0 , then maximum

The second derivative is.

f''(x)=12x^2-4

and f''(0)=-4<0rArr (0,1)" is a maximum"

f''(-1)=8>0rArr(-1,0)" is a minimum"

f''(1)=8>0rArr(1,0)" is a minimum"
graph{x^4-2x^2+1 [-10, 10, -5, 5]}