Finding Absolute Extrema — Question 6

Question 6 - Solution

We are given: $$f(x) = x^4 - 4x^2$$

(a) Find critical points:

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

Set derivative to zero: $$f'(x) = 0 \Rightarrow x = 0,\, \pm\sqrt{2}$$

(b) Evaluate at critical points and endpoints:

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

(c) Conclusion:

• Absolute maximum: $\boxed{f(-3) = f(3) = 45}$

• Absolute minimum: $\boxed{f(\pm \sqrt{2}) = -4}$

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