let\nf(x)=\begin{cases}-x + b, &\text{if }x < - 4\\2, &\text{if }x=-4\\\frac{6}{x - 5}-1, &\text{if }x >…

Explanation:

Step1: Recall left - hand limit

The left - hand limit as $x\to - 4^{-}$ is $\lim_{x\to - 4^{-}}f(x)=\lim_{x\to - 4^{-}}(-x + b)=4 + b$.

Step2: Recall right - hand limit

The right - hand limit as $x\to - 4^{+}$ is $\lim_{x\to - 4^{+}}f(x)=\lim_{x\to - 4^{+}}\left(\frac{6}{x - 5}-1\right)=\frac{6}{-4 - 5}-1=-\frac{6}{9}-1=-\frac{2}{3}-1=-\frac{5}{3}$. And $f(-4)=2$.

(a)

For $f(x)$ to be continuous at $x = - 4$, we need $\lim_{x\to - 4^{-}}f(x)=\lim_{x\to - 4^{+}}f(x)=f(-4)$. Since $\lim_{x\to - 4^{-}}f(x)=4 + b$, $\lim_{x\to - 4^{+}}f(x)=-\frac{5}{3}$ and $f(-4)=2$, there is no value of $b$ for which $4 + b=-\frac{5}{3}=2$. So there is no solution for $b$ and the answer is $\varnothing$.

(b)

A removable discontinuity occurs when $\lim_{x\to - 4^{-}}f(x)=\lim_{x\to - 4^{+}}f(x)\neq f(-4)$. Set $4 + b=-\frac{5}{3}$, then $b=-\frac{5}{3}-4=-\frac{5 + 12}{3}=-\frac{17}{3}$.

Answer:

(a) $\varnothing$ (b) $-\frac{17}{3}$ (c) There is no value of $b$ that will cause an infinite discontinuity. As $x\to - 4$, neither $-x + b$ nor $\frac{6}{x - 5}-1$ has a vertical - asymptote behavior at $x=-4$. So the answer is $\varnothing$. (d) A jump discontinuity occurs when $\lim_{x\to - 4^{-}}f(x)\neq\lim_{x\to - 4^{+}}f(x)$. We have $\lim_{x\to - 4^{-}}f(x)=4 + b$ and $\lim_{x\to - 4^{+}}f(x)=-\frac{5}{3}$. $4 + b\neq-\frac{5}{3}$, so $b\neq-\frac{5}{3}-4=-\frac{17}{3}$. In interval notation, $b\in(-\infty,-\frac{17}{3})\cup(-\frac{17}{3},\infty)$.