1. ## Find Delta's Algebraically

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

I set it up like this:

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

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?

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

Combine terms:

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

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

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

Stuck.

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

I set it up like this:

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

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?

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

Combine terms:

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

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

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

Stuck.
We need to show that

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

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

So,

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

Assuming $\displaystyle m>0$ we obtain

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