Finding the constants a and b to make continuous on entire real line

**hemi** · May 31st 2009, 07:45 PM

Given:

$f(x)=\left\{\begin{array}{cc}2,&\mbox{ if } x\leq -1\\ ax+b,& \mbox{ if } \--1 < x < 3\\ -2, & \mbox{ if } x\geq 3\end{array}\right.$

I need to find the constants a and b such that the function is continous on the entire real number line. how can i do this ? (step by step is helpful)

**pankaj** · May 31st 2009, 07:59 PM

Originally Posted by hemi

Given:

$f(x)=\left\{\begin{array}{cc}2,&\mbox{ if } x\leq -1\\ ax+b,& \mbox{ if } \--1 < x < 3\\ -2, & \mbox{ if } x\geq 3\end{array}\right.$

I need to find the constants a and b such that the function is continous on the entire real number line. how can i do this ? (step by step is helpful)

A function is said to be continuous at $x=a$ if $\lim_{h\to 0}f(a-h)=\lim_{h\to 0}f(a+h)=f(a)$

$\lim_{h\to 0}f(-1-h)=-2$

$\lim_{h\to 0}f(-1+h)=\lim_{h\to 0}(a(-1+h)+b)=b-a$

Therefore, $b-a=-2$

$\lim_{h\to 0}f(3-h)=\lim_{h\to 0}(a(3-h)+b)=3a+b$

$\lim_{h\to 0}f(3+h)=\lim_{h\to 0}(-2)=-2$

$ 3a+b=-2 $

Solving for $a$ and $b$ we get $a=0$ and $b=-2$

**Amer** · May 31st 2009, 09:33 PM

Originally Posted by pankaj

A function is said to be continuous at $x=a$ if $\lim_{h\to 0}f(a-h)=\lim_{h\to 0}f(a+h)=f(a)$

$\lim_{h\to 0}f(-1-h)=""-2""$

$\lim_{h\to 0}f(-1+h)=\lim_{h\to 0}(a(-1+h)+b)=b-a$

Therefore, $b-a=""-""2$

$\lim_{h\to 0}f(3-h)=\lim_{h\to 0}(a(3-h)+b)=3a+b$

$\lim_{h\to 0}f(3+h)=\lim_{h\to 0}(-2)=-2$

$ 3a+b=-2 $

Solving for $a$ and $b$ we get $a=0$ and $b=-2$

just a small thing pankaj b-a=2 not -2 since f(-1)=2

**pankaj** · May 31st 2009, 09:55 PM

Sorry. I goofed up

**hemi** · May 31st 2009, 09:56 PM

will this make the final answer of b = 2 then rather then -2?

**Amer** · May 31st 2009, 10:00 PM

b-a=2
3a+b=-2 ok

a-b=-2
3a+b=-2 find sum two equation

4a=-4
a=-1 so

b-(-1)=2
b+1=2
b=1

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