# Calculusfor Stationary Points

TheoehT
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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Staff Emeritus
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I think you found the derivative wrong, it should be;
$$f'(x) = \frac{10x}{3}(5x^2 + 8)^{\frac{-2}{3}}$$

Just to confirm, you are trying to find the stationary points?

Staff Emeritus
Gold Member
Oh and by the way don't forget that;
$$x^{\frac{2}{3}} = (\sqrt[3]{x})^2$$

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TheoehT
thanks for the fixup, I am trying to find the stationary points

Staff Emeritus
Gold Member
You can re-write f'(x) like this;
$$f'(x) = \frac{10x}{3(5x^2 + 8)^{\frac{2}{3}}}$$

Can you go from here?

TheoehT
square roots are even worse

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?

TheoehT
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

Staff Emeritus
Gold Member
TheoehT said:
square roots are even worse
hang on... if f(x) = 0 then doesn't 10x have to equal 0

You've got it 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.

Staff Emeritus
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$$\frac{10x}{3(5x^2 + 8)^{\frac{2}{3}}}= 0$$

What's the only possible way that it could equal zero?

TheoehT
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!

TheoehT
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

Staff Emeritus
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Nope your denomenator would be twelve. All you need to do now is find your y co-ordinate.

TheoehT
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.

TheoehT
I have (8+5(-1)^2)^3
= 2.235 +ve.

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

Staff Emeritus
Gold Member
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 $f''(x) > 0$ then it is a minimum point.
(2)If $f''(x) < 0$ then it is a maximum point.
(2)If $f''(x) = 0$ then you need to look a small $\Delta x$ either side of the stationary point to decide the nature of the point.

Hope this helps

TheoehT
you've helped me a great deal, that ends 3 days of worrying!

Thanks.

Staff Emeritus