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amy0799

  • one year ago

if f'(2) = lim h-->0 (cos(pi(2+h))-1)/h what is f(3)?

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  1. amy0799
    • one year ago
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    |dw:1440370564478:dw|

  2. amy0799
    • one year ago
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    I really need help on this

  3. zzr0ck3r
    • one year ago
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    \(\cos(3\pi)\)

  4. amy0799
    • one year ago
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    how'd u get that?

  5. zzr0ck3r
    • one year ago
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    the function is \(\cos(\pi x)\)

  6. zzr0ck3r
    • one year ago
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    \(f'(x) = \lim_{h\rightarrow 0}\dfrac{\cos(\pi(x+h))-\cos(\pi x)}{h}\) Now plug in \(2\) and \(\cos(2\pi)\) reduces to \(1\)

  7. amy0799
    • one year ago
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    can u show me the steps to how you got that equation?

  8. zzr0ck3r
    • one year ago
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    The definition of the derivative of \(f\) is \(f'(x) =\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\) So if we plug in \(2\) we get \(f'(2) =\lim_{h\rightarrow 0} \dfrac{f(2+h)-f(2)}{h}\). Compare this to what you had and you see that the function is \(\cos(\pi*x)\).

  9. amy0799
    • one year ago
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    so what do you do after that?

  10. zzr0ck3r
    • one year ago
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    well, since we figured out the function is \(\cos(\pi x)\) we simply plug in \(3\) and get \[\cos(3\pi)=-1\]

  11. amy0799
    • one year ago
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    what do you do with the h in the denominator?

  12. zzr0ck3r
    • one year ago
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    Nothing, we don't need to actually run the limit, this question is not about the derivative, but the function itself.

  13. zzr0ck3r
    • one year ago
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    we looked at the derivative (the thing with the h) to get the function that we needed.

  14. amy0799
    • one year ago
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    ooh ok. why is the answer -1?

  15. zzr0ck3r
    • one year ago
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    \(\cos(3\pi)=-1\)

  16. zzr0ck3r
    • one year ago
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    It asks for \(f(3)\) not \(f'(3)\)

  17. amy0799
    • one year ago
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    I'm not understanding why it's negative

  18. anonymous
    • one year ago
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    you lost me there

  19. amy0799
    • one year ago
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    what do you mean?

  20. anonymous
    • one year ago
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    nvm i didn't read the question correctly i thought you were supposed to find the limit

  21. amy0799
    • one year ago
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    can you help me with another question?

  22. zzr0ck3r
    • one year ago
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    Do you understand this one?

  23. anonymous
    • one year ago
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    i guess you are suppose to think that if \[f'(2)=\lim_{h\to 0}\frac{\cos(\pi(2+h))-1}{h}\] then \[f(x)=\sin(\pi x)\]

  24. anonymous
    • one year ago
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    i don't

  25. zzr0ck3r
    • one year ago
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    you mean \(\cos(\pi x)\)?

  26. anonymous
    • one year ago
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    yeah that

  27. zzr0ck3r
    • one year ago
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    what are you confused about @satellite73 ? seems pretty basic intro calc question to me.

  28. zzr0ck3r
    • one year ago
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    maybe I am confused lol

  29. anonymous
    • one year ago
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    knowing the derivative does not mean you know the function

  30. amy0799
    • one year ago
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    if f(x)=2+6x-x^2, for what values of x is f'(x) positive?

  31. zzr0ck3r
    • one year ago
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    \((\cos(\pi *2))'=\lim_{h\rightarrow 0}\dfrac{\cos(\pi(2+h))-\cos(\pi*2)}{h}\implies f(x) = \cos(\pi x)\)

  32. anonymous
    • one year ago
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    lol

  33. zzr0ck3r
    • one year ago
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    hehe, I cant think of what else they are asking...

  34. anonymous
    • one year ago
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    that "plus k" is a killer

  35. amy0799
    • one year ago
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    can u guys help me with my other question?

  36. anonymous
    • one year ago
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    \[f(x)=2+6x-x^2\\ f'(x)=6-2x\] solve \[6-2x>0\]

  37. amy0799
    • one year ago
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    x<3?

  38. anonymous
    • one year ago
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    yup

  39. amy0799
    • one year ago
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    thank you! :)

  40. anonymous
    • one year ago
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    yw @zzr0ck3r i think the first question was just ill posed i am sure you answer was what they wanted \(\cos(3x)\)

  41. zzr0ck3r
    • one year ago
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    yeah , I didn't think of the +k thing, but I am not sure if they introduce them to that problem in intro differentiable clac. Honestly I forgot about the k thing lol. I hate calculus :)

  42. zzr0ck3r
    • one year ago
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    lol rereading my own remark makes me cringe

  43. anonymous
    • one year ago
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    it is not you, it is whoever wrote the stupid question

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