use the function and graph to answer the questions about existence, limits, and continuity.\nf(x) =…
use the function and graph to answer the questions about existence, limits, and continuity.\nf(x) = \\begin{cases}-x - 3, & - 5\\leq x < - 4 \\\\x + 5, & - 4 < x < - 2 \\\\-\frac{3}{2}x, & - 2 < x < 0 \\\\0, & 0\\leq x\\leq1 \\\\2x - 2, & 1 < x < 3 \\\\2, & x = 3 \\\\-2x + 10, & 3 < x < 5 \\\\1, & x = 5\\end{cases}\ndoes f(-2) exist?\nyes\nno\ndoes \\lim_{x\\to - 2}f(x) exist?\nno\nyes\ndoes \\lim_{x\\to - 2}f(x) equal f(-2)?\nno\nyes
Answer
Explanation:
Step1: Check if f(-2) exists
The function is defined as (f(x)=x + 5) for (-4<x<-2) and (f(x)=-\frac{3}{2}x) for (-2<x<0). There is no definition of the function at (x = - 2), so (f(-2)) does not exist.
Step2: Check if (\lim_{x\rightarrow - 2}f(x)) exists
We find the left - hand limit (\lim_{x\rightarrow - 2^{-}}f(x)) and right - hand limit (\lim_{x\rightarrow - 2^{+}}f(x)). For the left - hand limit, as (x\rightarrow - 2^{-}), (f(x)=x + 5), so (\lim_{x\rightarrow - 2^{-}}f(x)=-2 + 5=3). For the right - hand limit, as (x\rightarrow - 2^{+}), (f(x)=-\frac{3}{2}x), so (\lim_{x\rightarrow - 2^{+}}f(x)=-\frac{3}{2}\times(-2)=3). Since (\lim_{x\rightarrow - 2^{-}}f(x)=\lim_{x\rightarrow - 2^{+}}f(x)=3), (\lim_{x\rightarrow - 2}f(x)) exists.
Step3: Check if (\lim_{x\rightarrow - 2}f(x)=f(-2))
Since (f(-2)) does not exist and (\lim_{x\rightarrow - 2}f(x)=3), (\lim_{x\rightarrow - 2}f(x)\neq f(-2)) (because (f(-2)) is undefined).
Answer:
Does (f(-2)) exist? No Does (\lim_{x\rightarrow - 2}f(x)) exist? Yes Does (\lim_{x\rightarrow - 2}f(x)) equal (f(-2))? No