Consider the function

$$f(x) = \begin{cases}2x -1 & if & x < -1\\x^2 + 1 & if & -1\leq x \leq 1\\x + 1 & if & x > 1.\end{cases}$$

Then

F(x)=

         $$\lim_{x \rightarrow -1^{-}}$$ (2x-1) = 2(-1)-1 = -3
         $$\lim_{x \rightarrow -1^{+}}$$ $$(x^{2} + 1)$$ = 1+1 = 2
Since the left-hand limit and right-hand limit are not equal,the function is discontinuous at -1.
           $$\lim_{x \rightarrow 1^{-}}$$ $$(x^{2} + 1)$$ = 1+1 = 2
          $$\lim_{x \rightarrow -1^{+}}$$ x+1 = 1+1 = 2
Since the left-hand limit and right-hand limit are equal, the function is continuous at 1.
So the function is continuous everywhere except at -1.

C is the correct answer.