Calculusfor Stationary Points

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  • #1
TheoehT
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I'm having trouble with this question, because the 1/3 power is making my life very difficult! I don't 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
TheoehT
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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
TheoehT
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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 can't 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
TheoehT
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however, I still don't 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
TheoehT
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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
TheoehT
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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
TheoehT
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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
TheoehT
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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
TheoehT
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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 a lot 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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