# Did something wrong in my calculations, but I don't know what.

• Sep 8th 2010, 07:27 AM
ImaCowOK
Did something wrong in my calculations, but I don't know what.
$\displaystyle \frac{\frac{5}{6(x+h)+3}-\frac{5}{6x+3}}{h}$

Should end up in the form $\displaystyle \frac{A}{(Bx+Ch+3)(Dx+3)}$

I got my answer in that form and the values are correct for A, B, C, and D. But when I use it to answer the rest of the questions I'm getting wrong answers. I think I did something wrong in the beginning when getting it in that form.

Can someone go through the steps and give me your answer to check with mine. I need to see what I did wrong.

A = -30
B = 6
C = 6
D = 6

Can you display your final answer in the form: $\displaystyle \frac{A}{(Bx+Ch+3)(Dx+3)}$
• Sep 8th 2010, 07:48 AM
Unknown008
It seems good to me...

I'll take the numerator first.

$\displaystyle \frac{5}{6(x+h) + 3} - \frac{5}{6x+3} = \frac{5}{6x+6h + 3} - \frac{5}{6x+3}$

$\displaystyle = \frac{5(6x+3) - 5(6x+6h+3)}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{(30x+15) - (30x+30h+15)}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{30x+15 - 30x-30h-15}{(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{-30h}{(6x+6h + 3)(6x+3)}$

Now, the new fraction:

$\displaystyle \frac{\frac{5}{6(x+h) + 3} - \frac{5}{6x+3}}{h} = \frac{\frac{-30h}{(6x+6h + 3)(6x+3)}}{h}$

$\displaystyle = \frac{-30h}{h(6x+6h + 3)(6x+3)}$

$\displaystyle = \frac{-30}{(6x+6h + 3)(6x+3)}$
• Sep 8th 2010, 07:51 AM
ImaCowOK
That's exactly what I did, but how did you get rid of the h in the denominator? You got an error on one of your final steps.
• Sep 8th 2010, 07:55 AM
Unknown008
I missed a '}' in my post. I corrected it.

Remember that:

$\displaystyle \frac{(\frac{a}{b})}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}$
• Sep 8th 2010, 08:08 AM
ImaCowOK
Okay, I remember now. But I have to then use it to find $\displaystyle \lim_{h \to 0}$. I keep getting that wrong.

The other parts like f'(2), and f'(3) I've gotten right. f'(1) apparently is wrong and I did it exactly the way I did the others.
• Sep 8th 2010, 08:12 AM
Unknown008
Ok, now I'm not sure what you are talking about. Seems beyond what I've learned...

But there is an 'h' in the function. As h tends to zero, 6h will tend to zero, and the function tends to:

$\displaystyle = \frac{-30}{(6x+ 3)(6x+3)}$
• Sep 8th 2010, 08:27 AM
ImaCowOK
f'(1) and such means the derivative of the function when x is 1, so plug 1 into the function wherever there is an x. Sorry if you haven't learned it yet and I confused you.

Thanks for the answers you really helped me out. I had the right idea, but I was unsure about how to enter it.
• Sep 8th 2010, 08:33 AM
Unknown008
Ok, I'm familiar with derivatives. It's the limit that I'm not familiar with.

Uh... do you differentiate with the h? Because, if we already take it as h tends to zero, then,

$\displaystyle f(x) = \frac{-30}{(6x+ 3)(6x+3)} = -30(6x + 3)^{-2}$

$\displaystyle f'(x) = 60(6x+3)^{-3} . 6 = \frac{360}{(6x+3)^3}$
• Sep 8th 2010, 09:09 AM
ImaCowOK
$\displaystyle \lim_{h \to 0}$ was $\displaystyle \frac{-30}{(6x+ 3)(6x+3)}$ just as you said.

You substitute 0 in $\displaystyle \frac{-30}{(6x+6h + 3)(6x+3)}$ for h which leaves you with $\displaystyle \frac{-30}{(6x+ 3)(6x+3)}$ which was the correct answer. I didn't need to go any further for this particular exercise.