Questions: Complete parts (a) through (c) for the following function. f(x)=-3x^3+6x^2-4x-4 (a) Find intervals where the function is increasing or decreasing, and determine any relative extrema. (b) Find intervals where the function is concave upward or concave downward, and determine any inflection points. (c) Graph the function, considering the domain, critical points, symmetry, relative extrema, regions where the function is increasing or decreasing, inflection points, regions where the function is concave upward or concave downward, intercepts where possible, and asymptotes where applicable.

The function is decreasing on \((-\infty, \infty)\) and increasing nowhere.

To determine concavity, find the second derivative of the function.

\[ f''(x) = \frac{d}{dx}(-9x^2 + 12x - 4) = -18x + 12 \]

Set the second derivative equal to zero to find potential inflection points.

\[ -18x + 12 = 0 \implies x = \frac{12}{18} = \frac{2}{3} \]

Test intervals around the potential inflection point \( x = \frac{2}{3} \).

• For \( x < \frac{2}{3} \), choose \( x = 0 \): \[ f''(0) = -18(0) + 12 = 12 \quad (\text{positive, so } f(x) \text{ is concave upward}) \]

• For \( x > \frac{2}{3} \), choose \( x = 1 \): \[ f''(1) = -18(1) + 12 = -6 \quad (\text{negative, so } f(x) \text{ is concave downward}) \]

The function is concave upward on \((-\infty, \frac{2}{3})\) and concave downward on \((\frac{2}{3}, \infty)\). There is an inflection point at \( x = \frac{2}{3} \).

{"axisType": 3, "coordSystem": {"xmin": -3, "xmax": 3, "ymin": -10, "ymax": 10}, "commands": ["y = -3x^3 + 6x^2 - 4x - 4"], "latex_expressions": ["$y = -3x^3 + 6x^2 - 4x - 4$"]}