- x3_drummerchick

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

- katieb

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- x3_drummerchick

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- nincompoop

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

- Jhannybean

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

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## More answers

- Jhannybean

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

- Jhannybean

sorry, typo.

- Jhannybean

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

- x3_drummerchick

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!!

- Jhannybean

Woot no problem :D

- nincompoop

thank you, beanerwoman

- x3_drummerchick

you should be a teacher

- Jhannybean

...

- nincompoop

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

- Jhannybean

HAHA

- Jhannybean

So let's finish this badboy.

- Jhannybean

What did you get as your LCD?

- x3_drummerchick

one sec

- nincompoop

don't finish me :(

- x3_drummerchick

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

- Jhannybean

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

- x3_drummerchick

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

- x3_drummerchick

oh wait nvm

- Jhannybean

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

- x3_drummerchick

yeah im lost again :(

- x3_drummerchick

buut doesnt the denominator go away?

- Jhannybean

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!

- nincompoop

show that it goes away

- Jhannybean

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

- x3_drummerchick

okay

- x3_drummerchick

is this right?

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- Jhannybean

I'llcompare it with mine :) one minute!

- x3_drummerchick

okie dokie

- x3_drummerchick

i think i did somehing wrong

- x3_drummerchick

then i distribute negative here? :

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- x3_drummerchick

then i cancel out terms?

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- Jhannybean

Yeah I found where I had messed up too haha

- nincompoop

perform the operation instead of using the word cancel

- Jhannybean

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

- Jhannybean

So now all of this is divided by h.

- x3_drummerchick

holy sheesh

- Jhannybean

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

- Jhannybean

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

- x3_drummerchick

why are we multiplying it through the denominator only?

- Jhannybean

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

- x3_drummerchick

oh wow i didnt know that

- Jhannybean

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

- Jhannybean

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

- x3_drummerchick

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

- Jhannybean

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

- x3_drummerchick

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

- Jhannybean

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

- nincompoop

jhan, that looks pretty

- Jhannybean

Lol ty.

- x3_drummerchick

i would have never gotten this...

- Jhannybean

More importantly, do you understand my steps?

- x3_drummerchick

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!!

- Jhannybean

Yeah that's important \(\checkmark\)

- x3_drummerchick

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)

- Jhannybean

yes, good job.

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