The Shape of a Graph, Part I — Question 7

Question 7 - Solution

We are given: $$f(x) = x^3 - 3x^2 - 9x + 5$$

(a) First Derivative:

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

Set $f'(x) = 0$ to find critical points:

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

Test Intervals:

$(-\infty, -1)$: pick $x = -2$: $f'(-2) = 3(-2)^2 - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0$

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

$(3, \infty)$: pick $x = 4$: $f'(4) = 3(16) - 6(4) - 9 = 48 - 24 - 9 = 15 > 0$

Conclusion: $$\begin{aligned} &f(x) \text{ is increasing on } (-\infty, -1) \cup (3, \infty) \\ &f(x) \text{ is decreasing on } (-1, 3) \end{aligned}$$

(b) Local Extrema:

At $x = -1$: changes from increasing to decreasing → local maximum

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

At $x = 3$: changes from decreasing to increasing → local minimum

$f(3) = 27 - 27 - 27 + 5 = -22$

Answer: $$\text{Local maximum at } (-1, 10), \quad \text{Local minimum at } (3, -22)$$

Graph of $f(x) = x^3 - 3x^2 - 9x + 5$: