## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.4 Limits and Continuity - Exercises - Page 67: 49

#### Answer

$f$ is right-continuous in $x=1$ $f$ is continuous in $x=2$

#### Work Step by Step

We are given the function: $f(x)=\begin{cases} x^2+3,\text{ for }x<1\\ 10-x,\text{ for }1\leq x\leq 2\\ 6x-x^2,\text{ for }x>2 \end{cases}$ The functions $x^2+3$, $10-x$ and $6x-x^2$ are continuous on each of the subintervals as they are polynomial functions. Compute the left hand and right hand limits at $x=1$ and $x=2$: $\displaystyle\lim_{x\rightarrow 1^{-}} (x^2+3)=1^2+3=4$ $f(1)=10-1=9$ $\displaystyle\lim_{x\rightarrow 1^{+}} (10-x)=10-1=9$ $\displaystyle\lim_{x\rightarrow 2^{-}} (10-x)=10-2=8$ $f(2)=10-2=8$ $\displaystyle\lim_{x\rightarrow 2^{+}} (6x-x^2)=6(2)-2^2=8$ Therefore we got: $\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=f(1)=9$ $\Rightarrow f$ is right-continuous in $x=1$ $\displaystyle\lim_{x\rightarrow 2^{-}} f(x)=f(2)=\displaystyle\lim_{x\rightarrow 2^{+}} f(x)=8$ $\Rightarrow f$ is continuous in $x=2$

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