The Shape of a Graph, Part I — Question 9

Question 9 - Solution

We are given: $$f(x) = x^3 - 6x^2 + 9x + 1$$

(a) Critical Points:

Compute the first derivative: $$f'(x) = 3x^2 - 12x + 9$$

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

Critical points: $x = 1$, $x = 3$

(b) Increasing/Decreasing Intervals:

Test sign of $f'(x)$ in each interval:

$(-\infty, 1)$: pick $x = 0$, $f'(0) = 9 > 0$

$(1, 3)$: pick $x = 2$, $f'(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 < 0$

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

Increasing: $(-\infty, 1) \cup (3, \infty)$ Decreasing: $(1, 3)$

(c) Local Extrema:

At $x = 1$: increasing to decreasing → local max $f(1) = 1 - 6 + 9 + 1 = 5$

At $x = 3$: decreasing to increasing → local min $f(3) = 27 - 54 + 27 + 1 = 1$

Conclusion: $$\text{Local maximum at } (1, 5), \quad \text{Local minimum at } (3, 1)$$

Graph of $f(x) = x^3 - 6x^2 + 9x + 1$: