How do you find the local max and min for #f(x) = x^4 - 2x^2 + 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"#
#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(-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]}
