Finding Absolute Extrema — Question 9

Question 9 - Solution

We are given: $$f(x) = x^3 - 3x^2 + 1$$

(a) Find critical points:

$$f'(x) = 3x^2 - 6x = 3x(x - 2)$$

Set derivative to zero: $$3x(x - 2) = 0 \Rightarrow x = 0,\, x = 2$$

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

$$f(0) = 0 - 0 + 1 = 1$$ $$f(2) = 8 - 12 + 1 = -3$$ $$f(3) = 27 - 27 + 1 = 1$$

(c) Absolute extrema:

• Absolute minimum: $\boxed{f(2) = -3}$

• Absolute maximum: $\boxed{f(0) = f(3) = 1}$

Graph of $f(x) = x^3 - 3x^2 + 1$ on $[0, 3]$: