Critical Points — Question 1

Question 1 - Solution

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

(a) Find $f'(x)$: $$f'(x) = 3x^2 - 12x + 9.$$

Set $f'(x) = 0$ to find critical points: $$3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0.$$

Critical points: $x = 1, 3$

(b) First Derivative Test:

• Choose test values around $x=1$ and $x=3$.

• Use $f'(x) = 3x^2 - 12x + 9$.

Evaluate the sign of $f'(x)$:

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

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

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

Conclusion:

• At $x=1$, $f'(x)$ changes from positive to negative, so $f$ has a local maximum.

• At $x=3$, $f'(x)$ changes from negative to positive, so $f$ has a local minimum.

Answer:

• Local maximum at $\boxed{x=1}$ with $f(1)=5$

• Local minimum at $\boxed{x=3}$ with $f(3)=1$

Graph of $f(x)$ with critical points marked