#### Science Forum

1. 30 Sep '08 14:11
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)?
2. 30 Sep '08 14:30 / 1 edit
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.
Baby Gauss
30 Sep '08 15:30 / 1 edit
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?
4.  Palynka
Upward Spiral
30 Sep '08 15:34 / 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.

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?
Baby Gauss
30 Sep '08 15:43
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.
6.  Palynka
Upward Spiral
30 Sep '08 15:48
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.
Baby Gauss
30 Sep '08 15:49
Originally posted by Palynka
Blonde moment it was then.
I have two or three a day.
8. 30 Sep '08 16:07 / 3 edits
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
9. 30 Sep '08 16:43
f(6) is a number, so f'(6) is zero.
Baby Gauss
30 Sep '08 17:02 / 2 edits
Originally posted by Best101

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
Baby Gauss
30 Sep '08 17:05
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.
12. 30 Sep '08 17:41
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.
Baby Gauss
30 Sep '08 17:52
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.
14. 30 Sep '08 18:13
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.