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Calculusfor Stationary Points

  1. Mar 19, 2006 #1
    I'm having trouble with this question, because the 1/3 power is making my life very difficult! I dont have class for the next couple of days and without a teacher to ask, the question's driving me crazy!

    the question is find the stationary points for y = (8 +5x^2)^1/3

    using algebraic calculus

    I was able to find the derivative,
    dy/dx = 1/3(8+5x^2)^-2/3 times 10x
    = 10/3(8+5x^2)^-2/3

    = 10x (above)
    3(8+5x^2)^2/3

    I also know to make dy/dx = 0, but when I try to work it out I somehow get to 392^3/2 + 15x^11/2 - 10x = 0

    I have no idea how I got there! I'm really confused over the fractional power, and no examples I have with me deal with it!

    If anyone could help in some way, I'd really appreciate it.

    Thanks.
     
    Last edited: Mar 19, 2006
  2. jcsd
  3. Mar 19, 2006 #2

    Hootenanny

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    I think you found the derivative wrong, it should be;
    [tex]f'(x) = \frac{10x}{3}(5x^2 + 8)^{\frac{-2}{3}}[/tex]

    Just to confirm, you are trying to find the stationary points?
     
  4. Mar 19, 2006 #3

    Hootenanny

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    Oh and by the way don't forget that;
    [tex]x^{\frac{2}{3}} = (\sqrt[3]{x})^2[/tex]
     
    Last edited: Mar 19, 2006
  5. Mar 19, 2006 #4
    thanks for the fixup, I am trying to find the stationary points
     
  6. Mar 19, 2006 #5

    Hootenanny

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    You can re-write f'(x) like this;
    [tex]f'(x) = \frac{10x}{3(5x^2 + 8)^{\frac{2}{3}}}[/tex]

    Can you go from here? :smile:
     
  7. Mar 19, 2006 #6
    square roots are even worse :frown:

    hang on... if f`(x) = 0 then doesn't 10x have to equal 0, because you cant have the denominator equalling zero because it is undefined?

    I'm not sure if that is right, perhaps I have to move the denominator over the other side and try and solve?
     
  8. Mar 19, 2006 #7
    however, I still dont see how that gives me a stationary point, I've tried to bring the denominator over and it just gets into a mess!

    my attempts at this problem go over 5A4 sheets! and I still haven't got it
     
  9. Mar 19, 2006 #8

    Hootenanny

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    You've got it :smile: Its always a good idea to sketch the graph of your function first. So now all you have to do is find your y-co ordinate.
     
  10. Mar 19, 2006 #9

    Hootenanny

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    Just look at your differential;

    [tex]\frac{10x}{3(5x^2 + 8)^{\frac{2}{3}}}= 0 [/tex]

    What's the only possible way that it could equal zero?
     
  11. Mar 19, 2006 #10
    o.o

    all that time the answer was right under my nose. I could cry =p

    I'll go work graph the thing now, Thanks so much!
     
  12. Mar 19, 2006 #11
    with x = 0, because then 10x would = 0 and the denominator would still be a constant (albeit a messy one)

    therefore not undefined, I hope
     
  13. Mar 19, 2006 #12

    Hootenanny

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    Nope your denomenator would be twelve. All you need to do now is find your y co-ordinate. :smile:
     
  14. Mar 19, 2006 #13
    I took the extra step and worked out x= 0 or x= 12 an rejected x = 12.

    then, sub'ing the x value back into the equation I got
    8⅓ (got my head around the fraction, yay)
    3√8
    y=2.

    I'm on to graphing it, but aside from checking my calculator, how do I check the -nature- of stationary points? does that mean if it's a minimum or maximum? I seem to remember checking either side of the stationary point, but i'm not sure.
     
  15. Mar 19, 2006 #14
    I have (8+5(-1)^2)^3
    = 2.235 +ve.

    and that'll help me with my graph, hope it's right!
     
  16. Mar 19, 2006 #15

    Hootenanny

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    Yes, 'nature' means find whether it is a minimum, maximum or inflection, there is a simpler way to define the nature of the point. You need to find the second derivative;

    (1)If [itex]f''(x) > 0[/itex] then it is a minimum point.
    (2)If [itex]f''(x) < 0[/itex] then it is a maximum point.
    (2)If [itex]f''(x) = 0[/itex] then you need to look a small [itex]\Delta x[/itex] either side of the stationary point to decide the nature of the point.

    Hope this helps :smile:
     
  17. Mar 19, 2006 #16
    you've helped me a great deal, that ends 3 days of worrying!

    Thanks.
     
  18. Mar 19, 2006 #17

    Hootenanny

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    That look's right to me. Do you not have any plotting software? It would make this alot easier. If you haven't you can goto http://www.univie.ac.at/future.media/moe/onlinewerkzeuge.html to use a simple online graphing tool. :smile:
     
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