Calculusfor Stationary Points

  • Thread starter TheoehT
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  • #1
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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.
 
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Answers and Replies

  • #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?
 
  • #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]
 
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  • #4
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thanks for the fixup, I am trying to find the stationary points
 
  • #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:
 
  • #6
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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?
 
  • #7
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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
 
  • #8
Hootenanny
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TheoehT said:
square roots are even worse :frown:
hang on... if f`(x) = 0 then doesn't 10x have to equal 0

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.
 
  • #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?
 
  • #10
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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!
 
  • #11
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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
 
  • #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:
 
  • #13
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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.
 
  • #14
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I have (8+5(-1)^2)^3
= 2.235 +ve.

and that'll help me with my graph, hope it's right!
 
  • #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:
 
  • #16
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you've helped me a great deal, that ends 3 days of worrying!

Thanks.
 
  • #17
Hootenanny
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TheoehT said:
I have (8+5(-1)^2)^3
= 2.235 +ve.

and that'll help me with my graph, hope it's right!

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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