use the graph of function below to state in what interval(s) the function is increasing and decreasing…

Explanation:

Step1: Find the derivative of the function

$$f'(x) = 3x^2 - 8x$$

Step2: Determine critical points by setting derivative to zero

$$3x^2 - 8x = 0 \implies x(3x - 8) = 0 \implies x = 0, x = \frac{8}{3} \approx 2.67$$

Step3: Identify increasing and decreasing intervals

$$f'(x) > 0 \text{ on } (-\infty, 0) \cup (\frac{8}{3}, \infty); f'(x) < 0 \text{ on } (0, \frac{8}{3})$$

Step4: Calculate local maximum and minimum values

$$f(0) = 2; f(\frac{8}{3}) = (\frac{8}{3})^3 - 4(\frac{8}{3})^2 + 2 = \frac{512}{27} - \frac{256}{9} + 2 = -\frac{202}{27} \approx -7.48$$

Answer:

Increasing interval: $(-\infty, 0) \cup (\frac{8}{3}, \infty)$ Decreasing interval: $(0, \frac{8}{3})$ Local maximum: $(0, 2)$ Local minimum: $(\frac{8}{3}, -\frac{202}{27})$ or approximately $(2.67, -7.48)$