1. ## Continuity: finding b

Hi Community,

been trying to work out this continuity problem to find b. Could someone please take a look at my work?

Code:
Problem:
if f(x) is a function that is continuous then we must have b = ?

piecewise:

f(x) = 8x - 4 , x <= 5
-3x - b , x > 5
My work...

Code:
5 = f(5-) = f(5+)
5 = -3(5) + b
20 != b
Thank you!

2. $f(x)=\begin{cases}8x-4, \mbox{ if } x \le 5 \\ -3x-b, \mbox{ if } x > 5 \end{cases}$

$\lim_{x \to 5^-}8x-4=36$

Now

$\lim_{x \to 5^+}-3x-b=-15-b$

Now to be continous the above must be equal so

$36=-15-b \iff b=-51$

so we get

$f(x)=\begin{cases}8x-4, \mbox{ if } x \le 5 \\ -3x+51, \mbox{ if } x > 5 \end{cases}$

3. Originally Posted by mant1s
Hi Community,

been trying to work out this continuity problem to find b. Could someone please take a look at my work?

Code:
Problem:
if f(x) is a function that is continuous then we must have b = ?

piecewise:

f(x) = 8x - 4 , x <= 5
-3x - b , x > 5
My work...

Code:
5 = f(5-) = f(5+)
5 = -3(5) + b
20 != b
Thank you!
For a funnction to be continous at x=c, we must have

$\lim_{x\to{c^-}}f(x)=\lim_{x\to{c^+}}f(x)$

Therfore

$\lim_{x\to{5^-}}(8x-4)=\lim_{x\to{5^+}}(-3x-b)$

$\Longleftrightarrow{8(5)-4}=-3(5)-b$

You can do the rest