Critical Points — Question 9

Question 9 - Solution

(a) Find critical points:

Differentiate: $$f'(x) = 3x^2 - 6x - 9$$

Set $f'(x) = 0$: $$3x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0 \Rightarrow x = -1, \, 3$$

Critical points: $\boxed{x = -1, \, x = 3}$

(b) First Derivative Test:

Interval $(-\infty, -1)$: Choose $x = -2$, $f'(-2) = 3(4) + 12 - 9 = 15 > 0$

Interval $(-1, 3)$: Choose $x = 0$, $f'(0) = -9 < 0$

Interval $(3, \infty)$: Choose $x = 4$, $f'(4) = 3(16) - 24 - 9 = 15 > 0$

So: At $x = -1$, $f'$ changes from positive to negative → local maximum At $x = 3$, $f'$ changes from negative to positive → local minimum

(c) Function values:

$$f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$$ $$f(3) = 27 - 27 - 27 + 5 = -22$$

Local maximum: $\boxed{f(-1) = 10}$ Local minimum: $\boxed{f(3) = -22}$

Graph of the function: