the graph of f is given.\n(a) on what open intervals is f increasing?\n(b) on what open intervals is f…

Explanation:

Step1: Recall the relationship between (f') and (f)

If (f'(x)>0), (f(x)) is increasing. If (f'(x)<0), (f(x)) is decreasing. Local minima occur where (f') changes from negative to positive, and local maxima occur where (f') changes from positive to negative.

Step2: Analyze the graph of (f')

Looking at the graph, (f'(x)>0) on the intervals ((-\infty,-3)) and ((1,\infty)), so (f(x)) is increasing on these intervals. (f'(x)<0) on the interval ((-3,1)), so (f(x)) is decreasing on this interval. For local maxima, we check where (f') changes from positive to negative. At (x = - 3), (f') changes from positive (left of (x=-3)) to negative (right of (x = - 3)). At (x=1), (f') changes from negative (left of (x = 1)) to positive (right of (x=1)).

Answer:

(a) (f) is increasing on ((-\infty,-3)\cup(1,\infty)) (b) (f) is decreasing on ((-3,1)) (c) There is no local minima (since the problem only asks for local maxima in the given multiple - choice, but if we follow the rules: local minima would be where (f') changes from negative to positive. If we assume the full analysis, but for the given options in the problem (only about local maxima in the provided multiple - choice part), we focus on (d)) (d) The local maximum(s) is/are located at (x=-3) (assuming the graph analysis, if the multiple - choice in the problem has a wrong pre - filled value in the box. If we go by the correct derivative - function analysis for local maxima: when (f') changes sign from positive to negative. From the graph of (f'), at (x=-3), (f') changes from positive (left - hand side) to negative (right - hand side). So the answer for (d) is (x = - 3))