Minimum and Maximum Values — Question 1

Question 1 - Solution

We are given: $$f(x) = 2x^3 - 9x^2 + 12x + 1$$

(a) First derivative: $$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$

Critical points: $x = 1, 2$

(b) Use the second derivative: $$f''(x) = \frac{d}{dx}(f'(x)) = 12x - 18$$

Evaluate at each critical point: $$f''(1) = 12(1) - 18 = -6 \;\Rightarrow\; \text{local maximum at } x = 1$$ $$f''(2) = 12(2) - 18 = 6 \;\Rightarrow\; \text{local minimum at } x = 2$$

Answer: $$\text{Local maximum at } \boxed{x = 1}$$ $$\text{Local minimum at } \boxed{x = 2}$$

Graph of $f(x)$