The Shape of a Graph, Part II — Question 1

Question 1 - 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 $x = -1$ and $x = 3$.

For $x < -1$, the derivative is positive. For $-1 < x < 3$, the derivative is negative. For $x > 3$, the derivative is positive.

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

(b) Local extrema

At $x = -1$, the function changes from increasing to decreasing, so there is a local maximum. $$f(-1) = 10$$

At $x = 3$, the function changes from decreasing to increasing, so there is a local minimum. $$f(3) = -22$$

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

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

(c) Second derivative

$$f''(x) = 6x - 6$$

Setting $f''(x) = 0$ gives $x = 1$.

For $x < 1$, the second derivative is negative, so the graph is concave down. For $x > 1$, the second derivative is positive, so the graph is concave up.

The inflection point occurs at $$f(1) = -6$$

Concave down on $\boxed{(-\infty, 1)}$

Concave up on $\boxed{(1, \infty)}$

Inflection point at $\boxed{(1, -6)}$

Graph of $f(x)$