consider the function f given below.\n f(x)=\frac{x^{2}-4}{x + 2} \na) for what x - values(s) is this…

Explanation:

Step1: Simplify the function

First, factor the numerator $x^{2}-4=(x + 2)(x - 2)$. Then $f(x)=\frac{(x + 2)(x - 2)}{x+2}=x - 2$ for $x\neq - 2$. The function has a removable - discontinuity at $x=-2$. A function is not differentiable at points of discontinuity.

Step2: Analyze non - differentiable points

The function $y = f(x)$ is not differentiable at $x=-2$ because the original function is not defined at $x=-2$ (division by zero in the original form $\frac{x^{2}-4}{x + 2}$).

Step3: Find the derivative of the simplified function

Since $f(x)=x - 2$ for $x\neq - 2$, using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, the derivative $f'(x)=\frac{d}{dx}(x-2)=1$.

Step4: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $f'(x)$. Since $f'(x)=1$ for all $x\neq - 2$, then $f'(2)=1$.

Answer:

a) $-2$ b) $1$