Minimum and Maximum Values — Question 2

Question 2 - Solution

(a) Find critical points:

First, compute the derivative: $$f'(x) = 3x^2 - 12x + 9$$

Set the derivative to zero: $$3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0 \Rightarrow x = 1,\, x = 3$$

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

(b) Evaluate function at critical points and endpoints:

$$f(0) = 0 - 0 + 0 + 2 = 2$$ $$f(1) = 1 - 6 + 9 + 2 = 6$$ $$f(3) = 27 - 54 + 27 + 2 = 2$$ $$f(4) = 64 - 96 + 36 + 2 = 6$$

Conclusion:

- Maximum value: $\boxed{6}$ at $x = 1$ and $x = 4$ - Minimum value: $\boxed{2}$ at $x = 0$ and $x = 3$

Graph of the function on $[0, 4]$: