Question

The maximum value of the function \( f(x) = x^4 - 8x^2 + 2 \) for \( -2 \leq x \leq 2 \) is \(\underline{\hspace{2cm}}\).

Hint:

To find maximum or minimum values, first find the critical points by solving \( f'(x) = 0 \), then evaluate \( f(x) \) at those points.

Answer 1

Correct Answer: 2

To find the maximum value of the function, we first take the derivative of \( f(x) \):
\[ f'(x) = 4x^3 - 16x \] Now, set \( f'(x) = 0 \) to find the critical points:
\[ 4x^3 - 16x = 0 \implies 4x(x^2 - 4) = 0 \implies x = 0, \, \pm 2 \] Now, evaluate \( f(x) \) at \( x = -2, 0, 2 \):
\[ f(-2) = (-2)^4 - 8(-2)^2 + 2 = 16 - 32 + 2 = -14 \] \[ f(0) = 0^4 - 8(0)^2 + 2 = 2 \] \[ f(2) = (2)^4 - 8(2)^2 + 2 = 16 - 32 + 2 = -14 \] The maximum value occurs at \( x = 0 \), where \( f(0) = 2 \). Thus, the maximum value of the function is \( \boxed{2} \).