The given piecewise-defined function f(x),
[tex]f(x) = \begin{cases} 2x + 5 & \text{if } x \le -1 \\ -2x + 1 & \text{if } x \ge -1\end{cases}[/tex]
is continuous at x = -1 if both limits as x approaches -1 from either side are the same. We have
[tex]\displaystyle \lim_{x\to-1^-} f(x) = \lim_{x\to-1} (2x+5) = 3[/tex]
[tex]\displaystyle \lim_{x\to-1^+} f(x) = \lim_{x\to-1} (-2x+1) = 3[/tex]
so the claim is true, f(x) is indeed continuous at x = -1.
