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Weave
Jan29-07, 10:07 PM
1. The problem statement, all variables and given/known data
This problem is not hard at all its just that this stupid online homework program is a problem. Anyways,

f(x)=1/(5x+7)

The quotient:
f(7+h)+f(7)\h

This can be simplified to:
1\(A+Bh)

What is A & B
What is f'(7)?

2. Relevant equations

lim_{h\rightarrow0}\frac{\f(7+h)-f(7)}{h}


3. The attempt at a solution

The derivative is easy to get.
The final form is \frac{-5}{210h+1764}
How do I get it to:

[tex]\frac{1}{A+Bh}[\tex]

cristo
Jan29-07, 10:15 PM
Why not multiply top and bottom by -1/5?

Weave
Jan29-07, 10:21 PM
but when I input B=210 and A=1764 it tells me I'm incorrect

cristo
Jan29-07, 10:24 PM
Well, you haven't multiplied them by -1/5. In order to get \frac{-5}{210h+1764} into the required form, you need to multiply top and bottom by -1/5 (to get unity in the numerator).

Weave
Jan29-07, 10:37 PM
Ah..of coarse.
There is another problem that requires the form Ah^2 + B h + C.
f(x) = 2x^2 + 9 x + 4, find f'(2).

Now I got :
2h+17, A is 0, B is 2 but
when i input 17 for c it is incorrect. Why?

Gib Z
Jan29-07, 11:32 PM
Umm do you have to do it by first principles? It seems the quotient rule would work fine here.

For your 2nd problem, f'(2)=17, I dont see what the problem is.

Weave
Jan29-07, 11:38 PM
Umm do you have to do it by first principles? It seems the quotient rule would work fine here.

For your 2nd problem, f'(2)=17, I dont see what the problem is.

Well the question wants the form Ah^2+Bh+C
I know that
A=0
B=2
Shouldnt C=17? When i input that it is wrong.

Weave
Jan30-07, 12:19 AM
If not c=17, I don't understand what else it could be!