1. ## Find Delta's Algebraically

$f(x) = mx+b$ when $m>0$, $L = (m/2)+b$, X sub 0 = 1/2, $\epsilon = c>0$

I set it up like this:

$\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c>0$

Okay, so it looks like my restrictions are that it has to be greater than 0 for M and C. But I don't know how to deal with that in an inequality. Does that mean that I simply put C on both sides without one being a negative and a positive?

$c<\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c$

Combine terms:

$c<\arrowvert mx+\frac{m}{2}+2b\arrowvert< c$

What do I do here? I want X by itself? So.

$\frac{c}{m}<\arrowvert x+\frac{m}{2}+2b\arrowvert< \frac{c}{m}$

Stuck.

2. Originally Posted by A Beautiful Mind
$f(x) = mx+b$ when $m>0$, $L = (m/2)+b$, X sub 0 = 1/2, $\epsilon = c>0$

I set it up like this:

$\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c>0$

Okay, so it looks like my restrictions are that it has to be greater than 0 for M and C. But I don't know how to deal with that in an inequality. Does that mean that I simply put C on both sides without one being a negative and a positive?

$c<\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c$

Combine terms:

$c<\arrowvert mx+\frac{m}{2}+2b\arrowvert< c$

What do I do here? I want X by itself? So.

$\frac{c}{m}<\arrowvert x+\frac{m}{2}+2b\arrowvert< \frac{c}{m}$

Stuck.
We need to show that

$|f(x)-L|<\epsilon$ whenever $|x-a|<\delta$

And we have $a=\frac1{2}$ and $L=\frac{m}{2}+b$

So,

$|(mx+b)-(\frac{m}{2}+b)|=|m(x-\frac{1}{2})|<\epsilon$

Assuming $m>0$ we obtain

$|(x-\frac{1}{2})|<\frac{\epsilon}{m}=\delta$