I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.

Also, What exactly is the relationship of the definite integral to the derivative?

Again, help and or tips would be greatly appreciated.

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- Jan 6th 2009, 11:32 AMbemidjibasserHelp with delta epsilon proofs...
I need to prove that lim (mx+b) as x approaches c = mc +b. Any tips or help would be greatly appreciated.

Also, What exactly is the relationship of the definite integral to the derivative?

Again, help and or tips would be greatly appreciated. - Jan 6th 2009, 12:09 PMMoo
Hello,

For the second one, if f is a function and f ' is its derivative, we have :

$\displaystyle \int_a^b f'(x) ~dx=f(b)-f(a)$

Usually, we define an antiderivative of f, F and say $\displaystyle \int_a^b f(x) ~dx=F(b)-F(a)$

Since f is an antiderivative of f ', we have the formula above. - Jan 6th 2009, 12:12 PMgalactus
This actually is not that bad. I know that the epsilon-delta thing can be confusing.

$\displaystyle f(x)=mx+b, \;\ m\neq 0$

$\displaystyle \lim_{x\to c}(mx+b)=mc+b$

$\displaystyle |(mx+b)-(mc+b)|<{\epsilon}$

$\displaystyle |m(x-c)|<{\epsilon}$

$\displaystyle |x-c|<\frac{\epsilon}{|m|}={\delta}$

Therefore, for $\displaystyle {\epsilon}>0$, let $\displaystyle {\delta}=\frac{\epsilon}{|m|}$ - Jan 6th 2009, 03:41 PMbemidjibasser
Thank you both. Is there a way to expand on the relationship of the derivative and the definite intergal? I think I can follow the formulas above, but how would you go about explaining more in words? Thanks again for help.

- Jan 6th 2009, 05:02 PMMathstud28
An alternative to Moo's post would be that in a sense integration and differentiation are inverses. If $\displaystyle F(x)=\int_{a}^x f~dx$ and $\displaystyle f$ is continuous at some point $\displaystyle x_0$ then $\displaystyle F(x)$ is differentiable at $\displaystyle x_0$ and $\displaystyle F'(x_0)=f(x_0)$

- Jan 6th 2009, 05:35 PMbemidjibasser
Does the fundamental theorem have anything to do with this? Or the mean value theorem? Thank you again...

- Jan 6th 2009, 05:45 PMMathstud28