the figure below shows the graph of a function over the closed interval -1,2. answer parts (a) through (c)…

Explanation:

Step1: Recall differentiability and continuity concepts

A function is differentiable at a point if the left - hand and right - hand derivatives exist and are equal. A function is continuous at a point if the limit of the function as x approaches that point exists and is equal to the function value at that point.

Step2: Analyze part (a)

The function has a smooth curve (no sharp corners or cusps) on (-1 < x<0) and (0 < x\leq2). At (x = - 1) and (x=0) there are possible issues. At (x=-1) it is an endpoint and at (x = 0) there is a sharp corner. So the function is differentiable for (-1<x < 0,0 < x\leq2).

Step3: Analyze part (b)

At (x = 0), the function is continuous (the graph has no breaks) but has a sharp corner. So the left - hand and right - hand derivatives are not equal. Thus the function is continuous but not differentiable at (x = 0).

Step4: Analyze part (c)

The function appears to be continuous on the interval ([-1,2]) (no jumps or holes in the graph). So there are no points where the function is neither continuous nor differentiable.

Answer:

(a) A. (-1<x < 0,0 < x\leq2) (b) C. (x = 0) (c) D. None