- 30 Sep '08 15:30 / 1 edit

Are you sure you didn't mean to type F(6 + h) - F(6)= ah + bh^2 + ch^3 instead of f(6 + h) - F(6)= ah + bh^2 + ch^3. And maybe your question would be what's the value of F'(6). In this case F'(6)=f(6)=a.*Originally posted by Best101***Science is math so let's get down to business.**

I need help but first I need to see if you know what you are doing. Can anyone solve this problem before we move on?

1)Suppose thta f(x) is a function such that f(6 + h) - F(6)= ah + bh^2 + ch^3.

What's f '(6)?

Edit: And by saying you need help you aren't saying I need help to get my homewor done right? - 30 Sep '08 15:34 / 1 edit

Shouldn't it be zero?*Originally posted by adam warlock***Are you sure you didn't mean to type F(6 + h) - F(6)= ah + bh^2 + ch^3 instead of f(6 + h) - F(6)= ah + bh^2 + ch^3. And maybe your question would be what's the value of F'(6). In this case F'(6)=f(6)=a.**

Edit: And by saying you need help you aren't saying I need help to get my homewor done right?

If I take the limit definition of derivative. then since that expression is true \forall h, then it must also be when h = epsilon. It would then be the limit of a*epsilon +b*epsilon^2 + c *epsilon^3 when epsilon goes to zero, which is zero...

Or am I having a blonde moment? - 30 Sep '08 15:43

For you to have F'(6) you need to first divide by h and then take the limit when h goes to 0. When one divides by h the*Originally posted by Palynka***Shouldn't it be zero?**

If I take the limit definition of derivative. then since that expression is true \forall h, then it must also be when h = epsilon. It would then be the limit of a*epsilon +b*epsilon^2 + c *epsilon^3 when epsilon goes to zero, which is zero...

Or am I having a blonde moment?*a*term is h-less and so the limit is a. - 30 Sep '08 16:07 / 3 edits

This is actually the answer I had got but I wasn't too sure about it.*Originally posted by adam warlock***For you to have F'(6) you need to first divide by h and then take the limit when h goes to 0. When one divides by h the***a*term is h-less and so the limit is a.

On your last post I meant the first one with the two capital Fs (the lower case F was a typo). Also this is for study not homework :p

Edit: I think I just got stuck on another problem.

Directions: Compute f '(x) using the limit definition

F(x) = x/(x-1)

I got 0/0, but there's no way for me to check my answer. Help please - 30 Sep '08 17:02 / 2 edits

F(x+h)-F(x)=(x+h)/(x+h-1)-x/(x-1)=[(x+h)(x-1)-x(x+h-1)]/[(x+h-1)(x-1)]=*Originally posted by Best101***This is actually the answer I had got but I wasn't too sure about it.**

On your last post I meant the first one with the two capital Fs (the lower case F was a typo). Also this is for study not homework :p

Edit: I think I just got stuck on another problem.

Directions: Compute f '(x) using the limit definition

F(x) = x/(x-1)

I got 0/0, but there's no way for me to check my answer. Help please

=(x^2-x+hx-h-x^2-xh+x)/[(x+h-1)(x-1)]=-h/[(x+h-1)(x-1)]=

Now [F(x+h)-F(x)]/h=-1/[(x+h-1)(x-1)]

And taking the limit as h go to 0 we are left with F'(x)=-1/(x-1)^2 - 30 Sep '08 17:52

The OP already stated that he had made a typo so I guess you should rethink your stance.*Originally posted by Eladar***I'm saying you are wrong. This is a fundamental theorem of calculus problem. The function f(x) is being defined by an integral beginning at 6 and ending at 6+h.**

And I know calculus well enough to know I'm right. - 30 Sep '08 18:13The original problem was supposed to be:

F(6+h) - F(6) = ah + bh^2 + ch^3.

This means that the original problem was the integral from 6 to 6+h. This could be rewritten as the integral from 0 to 6+h minus the integral from 0 to 6. The resulting ah + bh^2+ch^3 is what's left over after subtracting the two integrals. In other words the equation represents the area under the curve from 6 to 6+h. I has nothing to do with what happens at 6. It has to do with the change over the interval 6 to 6+h.

I suppose we'll just have to wait until he gets the correct answer to see who is right. - 30 Sep '08 19:06

And he also wants to know F'(6) not f'(6). And even if he wanted to know f'(6) your argument to say it would equal 0 is wrong.*Originally posted by Eladar***The original problem was supposed to be:**

F(6+h) - F(6) = ah + bh^2 + ch^3.

This means that the original problem was the integral from 6 to 6+h. This could be rewritten as the integral from 0 to 6+h minus the integral from 0 to 6. The resulting ah + bh^2+ch^3 is what's left over after subtracting the two integrals. In other words the equation represen ...[text shortened]... +h.

I suppose we'll just have to wait until he gets the correct answer to see who is right.