Calculus Help

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)?

What you are loooking at is actually a second derivative problem. f'(6) = F''(6)

When I took the second derivative and solved for f'(6) I got:

f'(6) = f''(6+h) - 2b-6ch

Perhaps there's a substitution for f''(h+6), but I'm not seeing it.

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)?

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? 😛

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? 😛

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?

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?

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.

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.

Blonde moment it was then. 🙂

Originally posted by Palynka
Blonde moment it was then. 🙂

I have two or three a day. 🙂

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.

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 🙂

f(6) is a number, so f'(6) is zero.

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 🙂

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)]=
=(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

Originally posted by Eladar
f(6) is a number, so f'(6) is zero.

This is not the interpretation of f'(6).
If you want you can think you calculate f'(x) and then substitute x for 6 and calculate the result.

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.

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.

The OP already stated that he had made a typo so I guess you should rethink your stance.

And I know calculus well enough to know I'm right.

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 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.

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.

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.