Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

sorry, typo.

Answer 5

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

Answer 6

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!!

Answer 7

Woot no problem :D

Answer 8

thank you, beanerwoman

Answer 9

you should be a teacher

Answer 10

...

Answer 11

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

Answer 12

HAHA

Answer 13

So let's finish this badboy.

Answer 14

What did you get as your LCD?

Answer 15

one sec

Answer 16

don't finish me :(

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

oh wait nvm

Answer 21

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

Answer 22

yeah im lost again :(

Answer 23

buut doesnt the denominator go away?

Answer 24

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!

Answer 25

show that it goes away

Answer 26

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

Answer 27

okay

Answer 28

is this right?

Answer 29

I'llcompare it with mine :) one minute!

Answer 30

okie dokie

Answer 31

i think i did somehing wrong

Answer 32

then i distribute negative here? :

Answer 33

then i cancel out terms?

Answer 34

Yeah I found where I had messed up too haha

Answer 35

perform the operation instead of using the word cancel

Answer 36

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

Answer 37

So now all of this is divided by h.

Answer 38

holy sheesh

Answer 39

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

Answer 40

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

Answer 41

why are we multiplying it through the denominator only?

Answer 42

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

Answer 43

oh wow i didnt know that

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

jhan, that looks pretty

Answer 51

Lol ty.

Answer 52

i would have never gotten this...

Answer 53

More importantly, do you understand my steps?

Answer 54

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!!

Answer 55

Yeah that's important \(\checkmark\)

Answer 56

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)

Answer 57

yes, good job.