explain why the function is discontinuous at the given number a. (select all that apply.)\n…

explain why the function is discontinuous at the given number a. (select all that apply.)\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 apply.)\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) )

When ( x = -4 ), for the piece - wise function ( f(x)=\begin{cases}\frac{1}{x + 4}&x\neq - 4\1&x=-4\end{cases}), substituting ( x=-4 ) into ( \frac{1}{x + 4} ), we get ( \frac{1}{-4 + 4}=\frac{1}{0} ), which is undefined. And ( f(-4) = 1 ) is defined from the second part of the piece - wise function. But if we consider the first part (the non - constant part near ( x=-4 )), the value of the function from the non - constant part at ( x=-4 ) is undefined.

Step2: Calculate the left - hand limit

For the left - hand limit as ( x\rightarrow - 4^{-} ), ( f(x)=\frac{1}{x + 4} ). Let ( x=-4-h ), where ( h>0 ) and ( h\rightarrow0 ). Then ( \lim_{x\rightarrow - 4^{-}}f(x)=\lim_{h\rightarrow0}\frac{1}{-4 - h+4}=\lim_{h\rightarrow0}\frac{1}{-h}=-\infty )

Step3: Calculate the right - hand limit

For the right - hand limit as ( x\rightarrow - 4^{+} ), ( f(x)=\frac{1}{x + 4} ). Let ( x=-4 + h ), where ( h>0 ) and ( h\rightarrow0 ). Then ( \lim_{x\rightarrow - 4^{+}}f(x)=\lim_{h\rightarrow0}\frac{1}{-4 + h+4}=\lim_{h\rightarrow0}\frac{1}{h}=\infty )

Answer:

( f(-4) ) is undefined. ( \lim_{x\rightarrow - 4^{+}}f(x) ) and ( \lim_{x\rightarrow - 4^{-}}f(x) ) are not finite.