Science:Math Exam Resources/Courses/MATH104/December 2016/Question 09 (a)

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MATH104 December 2016

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Question 09 (a)
Consider the following function; $f(x)={\begin{cases}e^{x},&x\leq 0\\ax+b&0<x\leq 1\\1+{\sqrt {x}}&x>1.\end{cases}}$ (a) Determine values for $a$ and $b$ for which $f(x)$ is continuous.

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Hint
Recall the definition of the continuity. Find $a$ and $b$ to make $f$ continuous especially at $x=0$ and $x=1$ .

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Solution
Since each piece of functions $e^{x}$ , $ax+b$ , and $1+{\sqrt {x}}$ are continuous on its domain, it is enough to make $f$ continuous at $x=0$ and $x=1$ . Recall that $f$ is continuous at $x=c$ if $\lim _{x\to c^{+}}f(x)=\lim _{x\to c^{-}}f(x)=f(c)$ . First consider the continuity at $x=0$ . Since $\lim _{x\to 0^{-}}f(x)=\lim _{x\to 0^{-}}e^{x}=e^{0}=1=f(0)$ and $\lim _{x\to 0^{+}}f(x)=\lim _{x\to 0^{+}}ax+b=b$ , we need $b=1$ for the continuity. On the other hand, using $\lim _{x\to 1^{-}}f(x)=\lim _{x\to 1^{-}}ax+b=a+b=a+1=f(1)$ and $\lim _{x\to 1^{+}}f(x)=\lim _{x\to 1^{+}}1+{\sqrt {x}}=2$ , $a$ has to satisfy $a+1=2$ , i.e., $a=1$ to make $f$ continuous at $x=1$ . To sum, $\color {blue}a=1,b=1$ .