Finding Absolute Extrema — Question 2

Question 2 - Solution

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

(a) Find critical points:

Take the derivative: $$f'(x) = 3x^2 - 6x = 3x(x - 2)$$

Set $f'(x) = 0$: $$3x(x - 2) = 0 \Rightarrow x = 0,\, x = 2$$

Both are in the interval $[0, 4]$.

(b) Evaluate function at critical points and endpoints:

$$f(0) = 0^3 - 3(0)^2 + 1 = 1$$ $$f(2) = 8 - 12 + 1 = -3$$ $$f(4) = 64 - 48 + 1 = 17$$

(c) Absolute extrema:

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

• Absolute maximum: $\boxed{f(4) = 17}$

Graph of $f(x) = x^3 - 3x^2 + 1$ on $[0, 4]$: