The Shape of a Graph, Part I — Question 2

Question 2 - Solution

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

(a) First derivative

$$f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$$

The critical points are $$x = -1 \quad \text{and} \quad x = 3$$

We test the sign of $f'(x)$.

For $x < -1$, $f'(x) > 0$. For $-1 < x < 3$, $f'(x) < 0$. For $x > 3$, $f'(x) > 0$.

Therefore, the function is increasing on $\boxed{(-\infty, -1) \cup (3, \infty)}$

and decreasing on $\boxed{(-1, 3)}$.

(b) Local extrema

Since the function changes from increasing to decreasing at $x = -1$, there is a local maximum: $$f(-1) = 10$$

Since the function changes from decreasing to increasing at $x = 3$, there is a local minimum: $$f(3) = -22$$

Answer

Local maximum at $\boxed{(-1, 10)}$

Local minimum at $\boxed{(3, -22)}$

Graph of $f(x)$