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So I tried this out and got stuck with this:

$$0 = 3x^{(7/6)} + 2x - 10$$

I didn't think I could use a quadratic for this since its to the power of $7/6$

Here is the working I did:

We know its a critical point when f'(a) = 0

So I found the derivative of f(x) which is $$2*(5-x)/3x^{1/2} - x ^{2/3}$$

So I set this equal to 0

$$2*(5-x)/3x^{1/2} - x ^{2/3} = 0$$ $$2*(5-x)/3x^{1/2}=x ^{2/3}$$ $$2*(5-x)=x ^{2/3}\times3x^{1/2}$$ $$10-2x=3x ^{2/3 +1/2}$$ $$10=3x ^{7/6} + 2x$$

But this would be such a messy answer, so I think I have done something wrong with my working. Do you have any ideas?

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If $\ast$ means multiplication, your derivative is wrong. $x^{(2/3)}(5-x) = 5x^{2/3} - x^{5/3}$, and the derivative of that is $\frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3}$. – Daniel Fischer Oct 10 '13 at 13:18
up vote 2 down vote accepted

If $$f(x) = x^{2/3}(5-x) =5x^{2/3}-x^{5/3},$$

then $$f'(x) = \frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3}$$ Of course, to find critical points, we need to solve for $$f'(x) = 0 \iff \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{2/3}$$

Assuming if $x\neq 0$, multiply both sides of the equation by $\;\frac 35 x^{1/3}\;$ to find the solution. Note: Do check what happens at $x = 0$, where the function is defined, but not the derivative.

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Awesome, thank you. – Ghozt Oct 10 '13 at 13:37
You're welcome, Ghozt! – amWhy Oct 10 '13 at 13:37
Just one question, if we take f(0) = 0, what does this show? Is it a minima – Ghozt Oct 10 '13 at 13:39
$f(0) = 0$, but $f'(0)$ is undefined, since $$f'(x) = f'(x) = \frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3} = \frac{10}{3x^{1/3}} - \frac 53 x^{2/3}$$, and division by zero is undefined. So there is *no* maximum or minimum at $x = 0$ there, but it is interesting to note the behavior of $f'(x)$ as $x \to 0$. – amWhy Oct 10 '13 at 13:40
Ghozt: see for example the graph – amWhy Oct 10 '13 at 13:47

I'm not sure what happened when you were finding your derivative, but assuming that $*$ indicates multiplication, then it's wrong. $$x^{2/3}(5-x)=5x^{2/3}-x^{5/3}$$ Can you take it from there? (As a side note, critical points can also occur where the function is defined but the derivative is not.)

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Thank you, yes it is multiplication. I'm not sure what happened with the derivative :S – Ghozt Oct 10 '13 at 13:36

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