sabika13
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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imron07
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Can you find the first derivative of f(x)
sabika13
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for x=3.5, it is 14 and for x=20 is 2.36
imron07
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That's the slope of tangents at that points. (If you did find f'(x) correctly :))
sabika13
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the answer says -24 for 3.5 tho :S
imron07
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Tho? :? \[f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24 \]
sabika13
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though*.. I dont get it.. what did you do?
imron07
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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
mark_o.
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@ sabika13 did you get it? any question here?
mark_o.
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use the quotient rule here
mark_o.
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d(u/v)dx=(v du/dx - u dv/dx)/v^2
mark_o.
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can follow from here?
mark_o.
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for y= 2x/(x-3),
let u =2x and v=(x-3)
now do the quotient rule
mark_o.
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@ sabika13 are you following now here?
sabika13
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ivenever heard of the quotient rule.. or imforgetting..
sabika13
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what does d and v stand for :S
mark_o.
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oh i see, are you in algebra class now?
sabika13
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advanced functions
mark_o.
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yes, do you know limits?
mark_o.
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limits of a function?
mark_o.
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ok advance functions, im sure you already done limits of a function
sabika13
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im not sure if idid.. or what that is
mark_o.
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\[\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }\]
sabika13
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thanks
mark_o.
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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
mark_o.
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YW, good luck now