anonymous
  • anonymous
for the functions f(x)= 2x/(x-3), compare the slopes of the tangents i) at the points where x=3.5 and x=20 how do i do this?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Can you find the first derivative of f(x)
anonymous
  • anonymous
for x=3.5, it is 14 and for x=20 is 2.36
anonymous
  • anonymous
That's the slope of tangents at that points. (If you did find f'(x) correctly :))

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anonymous
  • anonymous
the answer says -24 for 3.5 tho :S
anonymous
  • anonymous
Tho? :? \[f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24 \]
anonymous
  • anonymous
though*.. I dont get it.. what did you do?
anonymous
  • anonymous
I thought it was some new math symbol. If f(x)=u/v then f'(x)=(u'v-uv')/v^2 Here, u=2x u'=2 v=x-3 v'=1
anonymous
  • anonymous
@ sabika13 did you get it? any question here?
anonymous
  • anonymous
use the quotient rule here
anonymous
  • anonymous
d(u/v)dx=(v du/dx - u dv/dx)/v^2
anonymous
  • anonymous
can follow from here?
anonymous
  • anonymous
for y= 2x/(x-3), let u =2x and v=(x-3) now do the quotient rule
anonymous
  • anonymous
@ sabika13 are you following now here?
anonymous
  • anonymous
ivenever heard of the quotient rule.. or imforgetting..
anonymous
  • anonymous
what does d and v stand for :S
anonymous
  • anonymous
oh i see, are you in algebra class now?
anonymous
  • anonymous
advanced functions
anonymous
  • anonymous
yes, do you know limits?
anonymous
  • anonymous
limits of a function?
anonymous
  • anonymous
ok advance functions, im sure you already done limits of a function
anonymous
  • anonymous
im not sure if idid.. or what that is
anonymous
  • anonymous
\[\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }\]
anonymous
  • anonymous
thanks
anonymous
  • anonymous
do that limit as x approach to zero and you will arrived at f'(x)=-6/(x-3)^2 then you plug in x=3.5 and x=20 that is f'(3.5)=-24 and f'(20)=-0.021 and for f(3.5)=14 and f(20)= 3.36
anonymous
  • anonymous
YW, good luck now

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