x3_drummerchick
  • x3_drummerchick
can someone explain to me how to do this problem? will give medals! (problem is in picture below)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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x3_drummerchick
  • x3_drummerchick
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nincompoop
  • nincompoop
your problem is very interesting how do you propose solving it if you were to figure it out?
Jhannybean
  • 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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Jhannybean
  • Jhannybean
So forget the \(h\) at the bottom for now and concentrate on finding an LCD between \(a+h+7\) and \(a-7\)
Jhannybean
  • Jhannybean
sorry, typo.
Jhannybean
  • Jhannybean
\[ \frac{f(a+h)-f(a)}{h} = \frac{\dfrac{4a+4h}{a+h-7} - \dfrac{4a}{a-7}}{h}\]
x3_drummerchick
  • 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
  • Jhannybean
Woot no problem :D
nincompoop
  • nincompoop
thank you, beanerwoman
x3_drummerchick
  • x3_drummerchick
you should be a teacher
Jhannybean
  • Jhannybean
...
nincompoop
  • nincompoop
she shouldn't like for real ;) she should be the CEO of openstudy
Jhannybean
  • Jhannybean
HAHA
Jhannybean
  • Jhannybean
So let's finish this badboy.
Jhannybean
  • Jhannybean
What did you get as your LCD?
x3_drummerchick
  • x3_drummerchick
one sec
nincompoop
  • nincompoop
don't finish me :(
x3_drummerchick
  • x3_drummerchick
cant you multiply whole thing by (a+h-7)and(a-7)? ?
Jhannybean
  • Jhannybean
Yup, That's what I got \[\frac{(a-7)(4a+4h)-4a(a+h-7)}{(a-7)(a+h-7)}\]
x3_drummerchick
  • x3_drummerchick
then it all cancels, well most of it, right?
x3_drummerchick
  • x3_drummerchick
oh wait nvm
Jhannybean
  • Jhannybean
No no, it doesn't all cancel. we have to expand it
x3_drummerchick
  • x3_drummerchick
yeah im lost again :(
x3_drummerchick
  • x3_drummerchick
buut doesnt the denominator go away?
Jhannybean
  • 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
  • nincompoop
show that it goes away
Jhannybean
  • 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
  • x3_drummerchick
okay
x3_drummerchick
  • x3_drummerchick
is this right?
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Jhannybean
  • Jhannybean
I'llcompare it with mine :) one minute!
x3_drummerchick
  • x3_drummerchick
okie dokie
x3_drummerchick
  • x3_drummerchick
i think i did somehing wrong
x3_drummerchick
  • x3_drummerchick
then i distribute negative here? :
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x3_drummerchick
  • x3_drummerchick
then i cancel out terms?
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Jhannybean
  • Jhannybean
Yeah I found where I had messed up too haha
nincompoop
  • nincompoop
perform the operation instead of using the word cancel
Jhannybean
  • 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
  • Jhannybean
So now all of this is divided by h.
x3_drummerchick
  • x3_drummerchick
holy sheesh
Jhannybean
  • Jhannybean
\[=\dfrac{\dfrac{-28h}{a^2+ah-14a-7h+49}}{h}\]
Jhannybean
  • Jhannybean
That means we're multiplying in h throughout the entire denominator.
x3_drummerchick
  • x3_drummerchick
why are we multiplying it through the denominator only?
Jhannybean
  • Jhannybean
Its a rule in dividing fractions: \(\dfrac{\dfrac{a}{b}}{c} \iff \dfrac{a}{bc} \)
x3_drummerchick
  • x3_drummerchick
oh wow i didnt know that
Jhannybean
  • Jhannybean
And actually, I just realized we don't eve have to multiply through the denominator, we can just leave it alone.
Jhannybean
  • 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
  • x3_drummerchick
so what do we do with the denominator now? thi is super intense 0_0
Jhannybean
  • 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
  • x3_drummerchick
holy rap girl. youre a genius... im gonna have to study this one.
Jhannybean
  • 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
  • nincompoop
jhan, that looks pretty
Jhannybean
  • Jhannybean
Lol ty.
x3_drummerchick
  • x3_drummerchick
i would have never gotten this...
Jhannybean
  • Jhannybean
More importantly, do you understand my steps?
x3_drummerchick
  • 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
  • Jhannybean
Yeah that's important \(\checkmark\)
x3_drummerchick
  • 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
  • Jhannybean
yes, good job.

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