Minimum and Maximum Values — Question 6

Question 6 - Solution

(a) Find critical points:

Differentiate: $$f'(x) = 4x^3 - 8x$$

Set derivative to zero: $$4x^3 - 8x = 0 \Rightarrow 4x(x^2 - 2) = 0 \Rightarrow x = 0, \pm\sqrt{2}$$

Critical points in $[-3, 3]$: $x = -\sqrt{2}, 0, \sqrt{2}$

(b) Evaluate $f(x)$ at critical points and endpoints:

$$f(-3) = (-3)^4 - 4(-3)^2 = 81 - 36 = 45$$ $$f(-\sqrt{2}) = (\sqrt{2})^4 - 4(\sqrt{2})^2 = 4 - 8 = -4$$ $$f(0) = 0^4 - 4(0)^2 = 0$$ $$f(\sqrt{2}) = 4 - 8 = -4$$ $$f(3) = 81 - 36 = 45$$

Conclusion:

- Absolute maximum: $\boxed{45}$ at $x = -3$ and $x = 3$ - Absolute minimum: $\boxed{-4}$ at $x = \pm\sqrt{2}$

Graph of $f(x) = x^4 - 4x^2$ on $[-3, 3]$: