1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
  2. jcsd
  3. Jan 29, 2005 #2

    $id

    User Avatar

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Maxima & Minima
Loading...