$$f(x) = \begin{cases} x+5 & x\lt -2 \\ x^2+2x+3 & x \geq -2 \end{cases}$$

Let $f$ be the function defined above. Which of the following statements about $f$ is true?

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We can check for continuity at $x=-2$ if the three hold true:

1. The limit at $x=-2$ exists.
2. $f(-2)$ exists.
3. The limit at $x=-2$ is equal to $f(-2)$.

The piecewise function is defined by a line and parabola. We first check that the limit exists by comparing the left and right hand limits.

Because $\lim\limits_{x\to-2^-}f(x) = \lim\limits_{x\to-2^+}f(x)$,  $\lim\limits_{x \to -2} = 3$.

$f(-2)$ can be found and is defined on the parabola.

$$f(-2)=(-2)^2+2(-2)+3$$ $$f(-2)=3$$

The limit and $f(-2)$ are equal, which means that $f$ is continuous at $x=-2$.

To check for differentiability, we use the same process but must find the derivative of $f$ first.

$$f(x) = \begin{cases} x+5 & x\lt -2 \\ x^2+2x+3 & x \geq -2 \end{cases}$$ $$f'(x)=\begin{cases} 1 & x\lt -2 \\ 2x+2 & x \geq -2 \end{cases}$$

Because the left and right hand limits of the derivative do not equal each other, $f$ is not differentiable at $x=-2$.