Question

the figure below shows the graph of a function over the closed interval -1,2. answer parts (a) through (c) (a) at what domain points does the function appear to be differentiable? a. -1≤x<0, 0<x≤2 b. x = 2 c. x=-1 d. none (b) at what domain points does the function appear to be continuous but not differentiable? a. x = 2 b. x=-1 c. x = 0 d. none

Explanation:

Step1: Recall differentiability condition

A function is differentiable where the graph has a smooth curve. At the endpoints of a closed - interval, we consider one - sided derivatives. At $x = 0$, the graph has a sharp corner, so it is not differentiable there.

Step2: Analyze domain for differentiability

For $-1\leq x<0$ and $0 < x\leq2$, the graph is a smooth curve. So the function is differentiable on $-1\leq x<0$ and $0 < x\leq2$.

Step3: Recall continuity and non - differentiability condition

A function is continuous if there are no breaks in the graph. A function is non - differentiable at a point if it has a sharp corner, cusp, or vertical tangent. At $x = 0$, the function is continuous (no breaks in the graph) but has a sharp corner, so it is not differentiable.

Answer:

(a) A. $-1\leq x<0,0 < x\leq2$
(b) C. $x = 0$