explain why the function is discontinuous at the given number a. (select all that ap\n f(x)=left{\begin{array…

explain why the function is discontinuous at the given number a. (select all that ap\n f(x)=left{\begin{array}{ll}\frac{1}{x + 4} & \text { if } x \neq-4 \\ 1 & \text { if } x=-4end{array} quad a=-4\right.\n( f(-4) ) is undefined.\n( lim _{x \rightarrow-4} f(x) ) is not finite.\n( f(-4) ) is defined and ( lim _{x \rightarrow-4} f(x) ) is finite, but they are not equal.\n( lim _{x \rightarrow-4^{+}} f(x) ) and ( lim _{x \rightarrow-4^{-}} f(x) ) are finite, but are not equal.\nnone of the above\nsketch the graph of the function.

explain why the function is discontinuous at the given number a. (select all that ap\n f(x)=left{\begin{array}{ll}\frac{1}{x + 4} & \text { if } x \neq-4 \\ 1 & \text { if } x=-4end{array} quad a=-4\right.\n( f(-4) ) is undefined.\n( lim _{x \rightarrow-4} f(x) ) is not finite.\n( f(-4) ) is defined and ( lim _{x \rightarrow-4} f(x) ) is finite, but they are not equal.\n( lim _{x \rightarrow-4^{+}} f(x) ) and ( lim _{x \rightarrow-4^{-}} f(x) ) are finite, but are not equal.\nnone of the above\nsketch the graph of the function.

Answer

Explanation:

Step1: Analyze ( f(-4) )

Given ( f(x)=\begin{cases}\frac{1}{x + 4}&x\neq - 4\1&x=-4\end{cases}), ( f(-4) = 1), so ( f(-4) ) is defined.

Step2: Calculate (\lim_{x\rightarrow - 4}f(x))

For ( x\neq - 4), ( f(x)=\frac{1}{x + 4}). As ( x\rightarrow - 4^{+}), let ( x=-4 + h), ( h\rightarrow0^{+}), then ( f(x)=\frac{1}{h}\rightarrow+\infty). As ( x\rightarrow - 4^{-}), let ( x=-4 - h), ( h\rightarrow0^{+}), then ( f(x)=\frac{1}{-h}\rightarrow-\infty). So (\lim_{x\rightarrow - 4}f(x)) does not exist (is not finite).

Answer:

(\lim_{x\rightarrow - 4}f(x)) is not finite.