f = x^4 - 8x^2 + 6
f' = 4x^3 - 16x
f" = 12x^2 - 16
f'=0 at x = 0,±2
at x = -2, f" > 0 so f has a local min
at x = 0, f" = < 0 so f has a local max
at x = 2, f" > 0 so f has a local min
So, now just evaluate f(x) at 0,±2 and the endpoints of the interval, to see what the absolute extrema are for the interval.
Find the absolute minimum and absolute maximum values of f on the interval below.
f(x) = x4 - 8x2 + 6
[-3, 4]
2 answers
good