17. which of the following describes the graph of y = |2x + 6|?\na. only continuous\nb. only…

Explanation:

Step1: Recall continuity definition

A function (y = f(x)) is continuous at a point (x = a) if (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)). For the absolute - value function (y = |2x + 6|), we can rewrite it as (y=\begin{cases}2x + 6, &\text{if }2x+6\geq0\text{ (i.e., }x\geq - 3)\-(2x + 6),&\text{if }2x+6<0\text{ (i.e., }x < - 3)\end{cases}). (\lim_{x\rightarrow - 3^{-}}|2x + 6|=\lim_{x\rightarrow - 3^{-}}-(2x + 6)=0), (\lim_{x\rightarrow - 3^{+}}|2x + 6|=\lim_{x\rightarrow - 3^{+}}(2x + 6)=0) and (|2(-3)+6| = 0). So (y = |2x + 6|) is continuous for all real - valued (x).

Step2: Recall differentiability definition

A function (y = f(x)) is differentiable at a point (x=a) if (\lim_{h\rightarrow0^{-}}\frac{f(a + h)-f(a)}{h}=\lim_{h\rightarrow0^{+}}\frac{f(a + h)-f(a)}{h}). For (x>-3), (y = 2x + 6), and (y^\prime=2). For (x < - 3), (y=-(2x + 6)), and (y^\prime=-2). (\lim_{h\rightarrow0^{-}}\frac{|2(-3 + h)+6|-|2(-3)+6|}{h}=\lim_{h\rightarrow0^{-}}\frac{|-6 + 2h+6|-0}{h}=\lim_{h\rightarrow0^{-}}\frac{|-2h|}{h}=\lim_{h\rightarrow0^{-}}\frac{-2h}{h}=-2) (\lim_{h\rightarrow0^{+}}\frac{|2(-3 + h)+6|-|2(-3)+6|}{h}=\lim_{h\rightarrow0^{+}}\frac{| - 6+2h + 6|-0}{h}=\lim_{h\rightarrow0^{+}}\frac{|2h|}{h}=\lim_{h\rightarrow0^{+}}\frac{2h}{h}=2) Since (\lim_{h\rightarrow0^{-}}\frac{|2(-3 + h)+6|-|2(-3)+6|}{h}\neq\lim_{h\rightarrow0^{+}}\frac{|2(-3 + h)+6|-|2(-3)+6|}{h}), the function (y = |2x + 6|) is not differentiable at (x=-3).

Answer:

a. only continuous