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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?

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Can you find the first derivative of f(x)
for x=3.5, it is 14 and for x=20 is 2.36
That's the slope of tangents at that points. (If you did find f'(x) correctly :))

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Other answers:

the answer says -24 for 3.5 tho :S
Tho? :? \[f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24 \]
though*.. I dont get it.. what did you do?
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
@ sabika13 did you get it? any question here?
use the quotient rule here
d(u/v)dx=(v du/dx - u dv/dx)/v^2
can follow from here?
for y= 2x/(x-3), let u =2x and v=(x-3) now do the quotient rule
@ sabika13 are you following now here?
ivenever heard of the quotient rule.. or imforgetting..
what does d and v stand for :S
oh i see, are you in algebra class now?
advanced functions
yes, do you know limits?
limits of a function?
ok advance functions, im sure you already done limits of a function
im not sure if idid.. or what that is
\[\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }\]
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
YW, good luck now

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