Make the function continuous everywhere

I'm sorry to come back with another Calculus question, to those of you who see me asking often.

I'm trying to make the following function continuous.

$\displaystyle f(x) = -1 $ if $\displaystyle x <= 0 $

$\displaystyle ax+b$ if $\displaystyle 0 < x < 1$

$\displaystyle 1 $ if $\displaystyle x >= 1$

I've seen simple examples which use limits to find the values but here I believe I would have to use the following limits, which will produce different values.

$\displaystyle \lim_{x \to -1} ax+b = -1$

$\displaystyle \lim_{x \to 1} ax+b = 1$

I honestly don't think this makes any sense.

Re: Make the function continuous everywhere

Quote:

Originally Posted by

**vaironxxrd** I'm trying to make the following function continuous.

$\displaystyle f(x) = -1 $ if $\displaystyle x <= 0 $

$\displaystyle ax+b$ if $\displaystyle 0 < x < 1$

$\displaystyle 1 $ if $\displaystyle x >= 1$

You need $\displaystyle {\lim _{x \to {0^ + }}}ax + b = - 1$ and $\displaystyle {\lim _{x \to {1^ - }}}ax + b = 1$

Re: Make the function continuous everywhere

Quote:

Originally Posted by

**Plato** You need $\displaystyle {\lim _{x \to {0^ + }}}ax + b = - 1$ and $\displaystyle {\lim _{x \to {1^ - }}}ax + b = 1$

But that means we would get two answers?

$\displaystyle \lim_{x \to 0^+} ax+b = -1$

=$\displaystyle a(0)+b = -1$

=$\displaystyle b = -1$

and

$\displaystyle \lim_{x \to 1^-}ax+b=1$

=$\displaystyle \lim_{x \to 1^-}a(1)+b=1$

=$\displaystyle \lim_{x \to 1^-}a+b=1$

I think I have to review some algebra 1 material (Crying)

Re: Make the function continuous everywhere

Quote:

Originally Posted by

**vaironxxrd** But that means we would get two answers?

$\displaystyle \lim_{x \to 0^+} ax+b = -1$

=$\displaystyle a(0)+b = -1$

=$\displaystyle b = -1$

and

$\displaystyle \lim_{x \to 1^-}ax+b=1$

Look you know that $\displaystyle b = -1$

so find the $\displaystyle a$ that makes $\displaystyle \lim_{x \to 1^-}ax-1=1$

Re: Make the function continuous everywhere

Quote:

Originally Posted by

**Plato** Look you know that $\displaystyle b = -1$

so find the $\displaystyle a$ that makes $\displaystyle \lim_{x \to 1^-}ax-1=1$

So to make this function continuous from 0<x<1...

b = -1

$\displaystyle \lim_{x \to 1-} ax-1 = 1$

= $\displaystyle a(1)-1 = 1$

= $\displaystyle a -1 = 1$

= $\displaystyle a = 2$

Therefore from x<0<1 the function is defined as ax+b = 2x -1

Re: Make the function continuous everywhere

Quote:

Originally Posted by

**vaironxxrd** So to make this function continuous from 0<x<1...

b = -1

$\displaystyle \lim_{x \to 1-} ax-1 = 1$

= $\displaystyle a(1)-1 = 1$

= $\displaystyle a -1 = 1$

= $\displaystyle a = 2$

Therefore from x<0<1 the function is defined as ax+b = 2x -1

$\displaystyle f(x)=\left\{ {\begin{array}{rc}{ - 1,}&{x < 0}\\{2x - 1,}&{0 \le x \le 1}\\{1,}&{1 < x}\end{array}} \right.$