Hello Marco,

The explanation that you give enough clarify the situation.

Cheers,

Rosa María

Hi Rosa,

If you mean the domain such that the last expression is positive and real, here is a Mathematica solution:

In[1]:= Reduce[5 x^(2/3) - x^(5/3) > 0]

Out[1]= 0 < x < 5

In[2]:= Plot[5 x^(2/3) - x^(5/3), {x, -5, 10}]

The Reduce expression also works in Alpha.

Kind regards,

David

Hi Rosa,

try to look up the difference between

x^(2/3)

and

Surd[x,3]^2

If in Mathematica you type

FunctionDomain[5 Surd[x, 3]^2 - Surd[x, 3]^5, x]

which evaluates to True, or use this wolfram alpha request you get what you expect. Note that for example

5 Surd[x, 3]^2 - Surd[x, 3]^5 /. x -> -4.

gives 22.6786 and

5 x^(2/3) - x^(5/3) /. x -> -4 // N

gives -11.3393 + 19.6402 I. Surd chooses the positive root. The "^" is for the Power function which calculates principal roots.

Cheers,

M.

Dear Rosa,

I am not really sure what your question is. Your two posts are not identical and the question seems to be in complete somehow.

In Mathematica / the Wolfram Language you would type:

FunctionDomain[5 x^(2/3) - x^(5/3), x]

Something similar works in Wolfram|Alpha, see here. I am not sure whether that helps.

Cheers,

M.