# 009A Sample Midterm 3, Problem 2 Detailed Solution

Sketch the graph of  ${\displaystyle f.}$  At each point of discontinuity, state whether  ${\displaystyle f}$  is left or right continuous.

${\displaystyle f(x)={\begin{array}{cc}{\Bigg \{}&{\begin{array}{cc}x^{3}+1&x\leq 0\\-x+1&0

Background Information:
Suppose  ${\displaystyle f(x)}$  is discontinuous at  ${\displaystyle x=a.}$
Now,  ${\displaystyle f}$  is left continuous at  ${\displaystyle x=a}$  if  ${\displaystyle \lim _{x\rightarrow a^{-}}f(x)=f(a).}$
Similarly,  ${\displaystyle f}$  is right continuous at  ${\displaystyle x=a}$  if  ${\displaystyle \lim _{x\rightarrow a^{+}}f(x)=f(a).}$

Solution:

Step 1:
We start by graphing  ${\displaystyle f.}$
Step 2:
From the graph, we can see that  ${\displaystyle f}$  is discontinuous at  ${\displaystyle x=2.}$
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 2^{+}}-x^{2}+10x-15}\\&&\\&=&\displaystyle {-4+20-15}\\&&\\&=&\displaystyle {1}\\&&\\&=&\displaystyle {f(2).}\end{array}}}$
Hence,  ${\displaystyle f}$  is right continuous at  ${\displaystyle x=2.}$

${\displaystyle f}$  is right continuous at  ${\displaystyle x=2.}$