use the function and the accompanying figure to answer the following questions.\na. is f defined at x =…

use the function and the accompanying figure to answer the following questions.\na. is f defined at x = 2?\nyes\nno\nb. is f continuous at x = 2?\nyes\nno\nf(x)=\\begin{cases}x^{2}-3, & -1\\leq x<0 \\\\ 4x, & 0<x<1 \\\\ 6, & x = 1 \\\\ -3x + 7, & 1<x<2 \\\\ 1, & 2\\leq x<3\\end{cases}

use the function and the accompanying figure to answer the following questions.\na. is f defined at x = 2?\nyes\nno\nb. is f continuous at x = 2?\nyes\nno\nf(x)=\\begin{cases}x^{2}-3, & -1\\leq x<0 \\\\ 4x, & 0<x<1 \\\\ 6, & x = 1 \\\\ -3x + 7, & 1<x<2 \\\\ 1, & 2\\leq x<3\\end{cases}

Answer

Explanation:

Step1: Check definition at x = 2

For (2\leq x<3), (f(x)=1). So (f(x)) is defined at (x = 2).

Step2: Check continuity at x = 2

Find left - hand limit (\lim_{x\rightarrow2^{-}}f(x)). For (1 < x<2), (f(x)=-3x + 7). Then (\lim_{x\rightarrow2^{-}}(-3x + 7)=-3\times2+7=1). Find right - hand limit (\lim_{x\rightarrow2^{+}}f(x)). For (2\leq x<3), (f(x)=1), so (\lim_{x\rightarrow2^{+}}f(x)=1). And (f(2) = 1). Since (\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x)=f(2)), the function is continuous at (x = 2).

Answer:

a. Yes b. Yes