question 35 (1 point) discuss the continuity for the function. be sure to include: a)whether it is…

Explanation:

Step1: Simplify the function

First, factor the numerator $x^{2}-4=(x + 2)(x - 2)$. So $f(x)=\frac{(x + 2)(x - 2)}{x+2}$. For $x\neq - 2$, we can cancel out the $(x + 2)$ terms, and $f(x)=x - 2$.

Step2: Check the limit as $x\to - 2$

We find $\lim_{x\to - 2}f(x)=\lim_{x\to - 2}(x - 2)$. Substitute $x=-2$ into $x - 2$, we get $\lim_{x\to - 2}(x - 2)=-2-2=-4$.

Step3: Check the value of the function at $x=-2$

The original function $f(x)=\frac{x^{2}-4}{x + 2}$ is not defined at $x=-2$ since the denominator is zero when $x=-2$.

Step4: Determine continuity and type of discontinuity

Since $\lim_{x\to - 2}f(x)=-4$ exists but $f(-2)$ is not defined, the function is discontinuous at $x =-2$. This is a removable - discontinuity because we can re - define the function $f(x)$ as $f(x)=\begin{cases}\frac{x^{2}-4}{x + 2},&x\neq - 2\-4,&x=-2\end{cases}$ to make it continuous.

Answer:

a) The function is discontinuous at $x=-2$. b) The type of discontinuity is removable discontinuity.