Odd/even function and critical points

[dyn]

Homework Statement

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

Identify domain of $f(x)$ and is it odd or even or neither ?
Find the critical points of f(x)

Relevant Equations

Even function f(x) = f(-x) , odd function f(x) = -f(-x)
Critical points have the 1st derivative equal to zero

I have $3x^{2/3}$ as an even function although there is some debate as to this in another thread I started but the (5-x) factor means the function is neither odd or even. I also see the domain as all real numbers. Hopefully this is right ?
To find the critical points I differentiate f(x) to get $10x^{-1/3}-5x^{2/3}$ and set this equal to zero to get the critical points.
I can get the critical point of x=2 from this. The answer also states that x=0 is a critical point. This is the bit that confuses me.
As it stands $10x^{-1/3}-5x^{2/3}$ is not defined at x=0 but if I rearrange it as $5x(2x^{-4/3}-x^{-1/3})$ I get both critical values of x=0 and x=2.
How can the same equation be defined and not defined at the same time ? Without knowing the answer how would I know to rearrange the equation to get x=0 as an answer ?
Thanks

 

[Mark44]

Homework Statement: $f(x) = 3x^{2/3}(5-x)$

Identify domain of $f(x)$ and is it odd or even or neither ?
Find the critical points of f(x)
Homework Equations: Even function f(x) = f(-x) , odd function f(x) = -f(-x)
Critical points have the 1st derivative equal to zero

I have $3x^{2/3}$ as an even function although there is some debate as to this in another thread I started

I don't think there is any debate about this being an even function.

but the (5-x) factor means the function is neither odd or even. I also see the domain as all real numbers. Hopefully this is right ?
To find the critical points I differentiate f(x) to get $10x^{-1/3}-5x^{2/3}$ and set this equal to zero to get the critical points.
I can get the critical point of x=2 from this. The answer also states that x=0 is a critical point. This is the bit that confuses me.
As it stands $10x^{-1/3}-5x^{2/3}$ is not defined at x=0 but if I rearrange it as $5x(2x^{-4/3}-x^{-1/3})$ I get both critical values of x=0 and x=2.

The second expression isn't defined at x = 0, either, as $x^{-1/3}$ isn't defined for this value.

How can the same equation be defined and not defined at the same time ? Without knowing the answer how would I know to rearrange the equation to get x=0 as an answer ?
Thanks

Critical points are where the derivative is 0, or at points in the domain of the function at which the derivative is undefined.

 

[dyn]

But if I re-arrange the equation I can get a function defined for all $x$ which gives $x=0$ as an answer

 

[Mark44]

But if I re-arrange the equation I can get a function defined for all $x$ which gives $x=0$ as an answer

No. I assume you are talking about this: $5x(2x^{-4/3}-x^{-1/3})$, which is undefined for x = 0. It's true that 5x = 0 when x = 0, but the other factor isn't defined, which makes the whole expression (it's not an equation) undefined.

Regarding what you wrote at the end of the other thread, $x^{1/3} = x^{2/6}$, obviously. The problem comes with the latter expression if you try to evaluate it as $\left( x^{1/6}\right)^2$ for negative x. What this shows, in my view, is that the usual rules for exponents don't carry through when you raise negative numbers to some fractional power.

 

[dyn]

Thank you Mark44 for your help. As I have previously mentioned in the other thread , it amazes me that the fractional indices for negative real numbers issue seems never mentioned in calculus of real numbers