Originally posted by Best101Are 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.
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)?
Originally posted by adam warlockShouldn't it be zero?
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 PalynkaFor 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.
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 adam warlockThis is actually the answer I had got but I wasn't too sure about it.
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 Best101F(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)]=
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
Originally posted by EladarThe OP already stated that he had made a typo so I guess you should rethink your stance.
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 EladarAnd 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.
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.