Use the second derivative test to find the relative extrema for f(x)=x^4-2x^2 + 3

If f(x) = x ⁴ - 2x ² + 3, compute the first and second derivatives:

f'(x) = 4x ³ - 4x

f''(x) = 12x ² - 4

Solve f'(x) = 0 to get the critical points of f(x) :

4x ³ - 4x = 4x (x ² - 1) = 0

→   x = -1, x = 0, x = 1

Check the sign of the second derivative at each of these points:

f'' (-1) = 8 > 0

f'' (0) = -4 < 0

f'' (1) = 8 > 0

These indicate the presence of

• two minima of f (-1) = 2 and f (1) = 2

• one maximum of f (0) = 3

which makes C the correct answer.

28

step-by-step explanation:

because that's the answer

on plase n