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Homework Help: Maxima & Minima

  1. Jan 29, 2005 #1
    Find the coordinates of a stationary point on the curve [tex]y= \frac{16x^3 + 4x^2 + 1}{2x^2}[/tex] And determine the nature of this points.

    How I find the coordinates? I know the second part.
     
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  3. Jan 29, 2005 #2

    $id

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    You have to find the dy/dx of that equation and find points where that is 0

    Hint: use the quotient rule to differentiate that
     
  4. Jan 29, 2005 #3

    arildno

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    You say you know about the second part.
    We are therefore, I hope, in agreement that the x-values for stationary points are found by the equation y'(x)=0.
    Let us denote a particular solution of this equation by X.
    But then, the corresponding y-value for a stationary point on the curve (for which the x-value is X) is simply y(X).
     
  5. Jan 29, 2005 #4
    Assuming that you mean this by a stationary point, set the first derivative equal to zero.

    EDIT--Oops...when I was replying to this post I didn't see others' responses. Anyway you have much better ones now. Cheers.
     
  6. Jan 29, 2005 #5
    How do you get the derivitive? When I let me derivitive equal to 0, I keep getting weird answers :grumpy:
     
  7. Jan 29, 2005 #6

    dextercioby

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    What do you mean by "weird"??Should they be "lovely"???

    On the other hand i don't see a really "nice" cubic...
    BTW,because "x=0" is not in the domain of the function,u can simplify the quartic on the numerator and end up with a cubic...

    Daniel.
     
  8. Jan 29, 2005 #7
    I did that the first time I did it. I got the derivitive [tex]8 - 4x^{-3}[/tex]
    Is that correct?
     
  9. Jan 29, 2005 #8

    dextercioby

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    Yes,it's correct.Now solve the equation into reals...

    Daniel.
     
  10. Jan 29, 2005 #9
    What else would he mean? :confused:
     
  11. Jan 29, 2005 #10
    When [tex]8 - 4x^{-3} = 0[/tex], I get 0.7937..... (thats why I said it was weird). I didn't anything wrong did I?
     
  12. Jan 29, 2005 #11

    dextercioby

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    If your answer is
    [tex] x=\frac{1}{\sqrt[3]{2}} [/tex]
    ,then it is correct.

    Daniel.
     
  13. Jan 29, 2005 #12
    Yeah, I got that. However my books answer is [tex](\frac{1}{2}, 8)[/tex]
     
  14. Jan 29, 2005 #13

    arildno

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    You have NOT calculated the derivative correctly (despite what others have told you):
    [tex]y'(x)=\frac{2x^{2}(3*16x^{2}+8x)-4x(16x^{3}+4x^{2}+1)}{4x^{4}}=\frac{2x}{4x^{4}}*((3*16x^{3}+8x^{2})-(2*16x^{3}+8x^{2}+2))=\frac{16x^{3}-2}{2x^{3}}=8-\frac{1}{x^{3}}[/tex]
    Hence, your root is [tex]X=\frac{1}{2}[/tex]
     
  15. Jan 29, 2005 #14

    arildno

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    Your book is right, BTW.
     
  16. Jan 29, 2005 #15

    dextercioby

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    U're right,Arildno,my mistake... :redface:

    The function,IIRC is
    [tex] y(x)=8x+4+\frac{1}{2x^{2}} [/tex]

    whose derivative is immediate
    [tex] y'(x)=8-x^{-3} [/tex]

    Daniel.
     
  17. Jan 29, 2005 #16

    arildno

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    footprints:
    Take this as a typical example of why I HATE the differentiation rule for fractions!
    It is the nastiest one, it is so easy to make a mistake.

    Daniel has kindly provided you with a rewriting which gives you the correct answer right away.
     
  18. Jan 29, 2005 #17
    I don't understand. When differentiating [tex] y(x)=8x+4+\frac{1}{2x^{2}} [/tex], won't I get [tex]8 - 4x^{-3}[/tex]?

    [tex]\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(2x^{-2-1})[/tex]
    [tex]\frac{dy}{dx}= 8 - 4x^{-3}[/tex]
    Whats wrong with that?
    Maybe you could show me in detail how it is done?
     
  19. Jan 29, 2005 #18

    arildno

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    This is wrong; it should be:

    [tex]\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(\frac{1}{2}x^{-2-1})[/tex]
     
  20. Jan 29, 2005 #19
    Oh right!!!! Thanks for the help guys!
     
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