009A Sample Midterm 3, Problem 2 Detailed Solution

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

$f(x)={\begin{array}{cc}{\Bigg \{}&{\begin{array}{cc}x^{3}+1&x\leq 0\\-x+1&0<x<2\\-x^{2}+10x-15&x\geq 2\end{array}}\end{array}}$

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

Solution:

Step 1:
We start by graphing  $f.$
Insert graph

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

Final Answer:
See Step 1 above for graph.
$f$  is right continuous at  $x=2.$

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