Question

Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.
f(x) = x4 - 8x² + 3

Answer 1

Answer:

The function is maxima at x =  0.

Step-by-step explanation:

the function is

$f(x) = x^4 - 8 x^2 + 3$

Differentiate with respect to x.

$f'(x) = 4x^3 - 16 x\\\\Put f'(x) = 0 \\\\4x(x^2-4)=0\\\\4 x(x +2)(x-2) =0\\\\x = 0, - 2 , 2 Now f''(x) = 12 x^2 - 16 \\\\f''(0) = - 16 = negative \\\\f''(-2) = 12(-2)^2 - 16 = 32\\\\f''(2)=12(2)^2 - 16 = 32$

So, the function is maima at x = 0 .