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sabika13

  • 3 years ago

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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  1. imron07
    • 3 years ago
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    Can you find the first derivative of f(x)

  2. sabika13
    • 3 years ago
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    for x=3.5, it is 14 and for x=20 is 2.36

  3. imron07
    • 3 years ago
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    That's the slope of tangents at that points. (If you did find f'(x) correctly :))

  4. sabika13
    • 3 years ago
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    the answer says -24 for 3.5 tho :S

  5. imron07
    • 3 years ago
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    Tho? :? \[f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24 \]

  6. sabika13
    • 3 years ago
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    though*.. I dont get it.. what did you do?

  7. imron07
    • 3 years ago
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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

  8. mark_o.
    • 3 years ago
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    @ sabika13 did you get it? any question here?

  9. mark_o.
    • 3 years ago
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    use the quotient rule here

  10. mark_o.
    • 3 years ago
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    d(u/v)dx=(v du/dx - u dv/dx)/v^2

  11. mark_o.
    • 3 years ago
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    can follow from here?

  12. mark_o.
    • 3 years ago
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    for y= 2x/(x-3), let u =2x and v=(x-3) now do the quotient rule

  13. mark_o.
    • 3 years ago
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    @ sabika13 are you following now here?

  14. sabika13
    • 3 years ago
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    ivenever heard of the quotient rule.. or imforgetting..

  15. sabika13
    • 3 years ago
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    what does d and v stand for :S

  16. mark_o.
    • 3 years ago
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    oh i see, are you in algebra class now?

  17. sabika13
    • 3 years ago
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    advanced functions

  18. mark_o.
    • 3 years ago
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    yes, do you know limits?

  19. mark_o.
    • 3 years ago
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    limits of a function?

  20. mark_o.
    • 3 years ago
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    ok advance functions, im sure you already done limits of a function

  21. sabika13
    • 3 years ago
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    im not sure if idid.. or what that is

  22. mark_o.
    • 3 years ago
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    \[\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }\]

  23. sabika13
    • 3 years ago
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    thanks

  24. mark_o.
    • 3 years ago
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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

  25. mark_o.
    • 3 years ago
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    YW, good luck now

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