## anonymous one year ago can someone explain to me how to do this problem? will give medals! (problem is in picture below)

1. anonymous

2. nincompoop

your problem is very interesting how do you propose solving it if you were to figure it out?

3. anonymous

ok so $f(a) = \frac{4a}{a-7}$$f(a+h) = \frac{4a+4h}{a+h+7}$ therefore... $\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h+7} - \dfrac{4a}{a-7}}{h}$

4. anonymous

So forget the $$h$$ at the bottom for now and concentrate on finding an LCD between $$a+h+7$$ and $$a-7$$

5. anonymous

sorry, typo.

6. anonymous

$\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h-7} - \dfrac{4a}{a-7}}{h}$

7. anonymous

ohh okay i see where this is going, the initial start confused me, i forgot which way to plug it in. oh and jhannybean. youre always to my rescue! you helped me understand things well enough yto get to precalc, so thanks!!

8. anonymous

Woot no problem :D

9. nincompoop

thank you, beanerwoman

10. anonymous

you should be a teacher

11. anonymous

...

12. nincompoop

she shouldn't like for real ;) she should be the CEO of openstudy

13. anonymous

HAHA

14. anonymous

15. anonymous

What did you get as your LCD?

16. anonymous

one sec

17. nincompoop

don't finish me :(

18. anonymous

cant you multiply whole thing by (a+h-7)and(a-7)? ?

19. anonymous

Yup, That's what I got $\frac{(a-7)(4a+4h)-4a(a+h-7)}{(a-7)(a+h-7)}$

20. anonymous

then it all cancels, well most of it, right?

21. anonymous

oh wait nvm

22. anonymous

No no, it doesn't all cancel. we have to expand it

23. anonymous

yeah im lost again :(

24. anonymous

buut doesnt the denominator go away?

25. anonymous

No, we're trying to find an LCD by multiplying both of the denominators of the two fractions with eachother, we can't simply cancel them otherwise there would be no point of finding the LCD in the first place!

26. nincompoop

show that it goes away

27. anonymous

And remember, we're still dividing EVERYTHING by h, so we need to expand our function to see which h values are still left :D

28. anonymous

okay

29. anonymous

is this right?

30. anonymous

I'llcompare it with mine :) one minute!

31. anonymous

okie dokie

32. anonymous

i think i did somehing wrong

33. anonymous

then i distribute negative here? :

34. anonymous

then i cancel out terms?

35. anonymous

Yeah I found where I had messed up too haha

36. nincompoop

perform the operation instead of using the word cancel

37. anonymous

$\frac{(a-7)(4a+4h)-4a(a+h-7)}{(a-7)(a+h-7)}$$=\frac{4a^2+4ah-28a-28h-4a^2-4ah+28a}{a^2+ah-7a-7a-7h+49}$$=\frac{\cancel{4a^2}\cancel{+4ah}\cancel{-28a}-28h\cancel{-4a^2}\cancel{-4ah}\cancel{+28a}}{a^2+ah-7a-7a-7h+49}$$=\frac{-28h}{a^2+ah-14a-7h+49}$

38. anonymous

So now all of this is divided by h.

39. anonymous

holy sheesh

40. anonymous

$=\dfrac{\dfrac{-28h}{a^2+ah-14a-7h+49}}{h}$

41. anonymous

That means we're multiplying in h throughout the entire denominator.

42. anonymous

why are we multiplying it through the denominator only?

43. anonymous

Its a rule in dividing fractions: $$\dfrac{\dfrac{a}{b}}{c} \iff \dfrac{a}{bc}$$

44. anonymous

oh wow i didnt know that

45. anonymous

And actually, I just realized we don't eve have to multiply through the denominator, we can just leave it alone.

46. anonymous

It would look like: $\dfrac{\dfrac{-28h}{a^2+ah-14a-7h+49}}{h} \longrightarrow \frac{-28h}{h(a^2+ah-14a-7h+49)}$

47. anonymous

so what do we do with the denominator now? thi is super intense 0_0

48. anonymous

And therefore, because we're multiplying both the numerator AND denominator by h, we can eliminate the h altogether. $\frac{-28\cancel{h}}{\cancel{h}(a^2+ah-14a-7h+49)} =\boxed{ -\frac{28}{a^2+ah-14a-7h+49}}$

49. anonymous

holy rap girl. youre a genius... im gonna have to study this one.

50. anonymous

So now we can state that $\boxed{\frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h-7} - \dfrac{4a}{a-7}}{h} = -\frac{28}{a^2+ah-14a-7h+49}}$

51. nincompoop

jhan, that looks pretty

52. anonymous

Lol ty.

53. anonymous

i would have never gotten this...

54. anonymous

More importantly, do you understand my steps?

55. anonymous

yes, and ive screenshotted all of them. im going to rewrite them all in my notes with your steps, i forgot about the division rule!!

56. anonymous

Yeah that's important $$\checkmark$$

57. anonymous

pretty much, we plugged in those functions in for x, got lcd common denominators, combined like terms, divided all by h in which we innstead, multiplied, then canceled out terms (h in num + denom)

58. anonymous

yes, good job.