Confused by results of FunctionDomain

Question

Why does Mathematica give me the function domain of $$f(x)=(2x-3)^{\frac{1}{5}}-(2x+1)^{\frac{1}{5}}-(x - 3)^{\frac{1}{5}}$$ as $x\geq3$?

I used this command:

FunctionDomain[(3 x - 2)^(1/5) - (2 x + 1)^(1/5) - (x - 3)^(1/5), x, Reals]

Answer 1

It has to do with how mathematica evaluates x^(1/5), which is in principle a multivalued function. In your case I assume you want the real root. You can do that using Surd. The following evaluates to True.

FunctionDomain[ Surd[3 x - 2, 5] - Surd[2 x + 1, 5] - Surd[(x - 3), 5], x, Reals]

Answer 2

At $x = 0$, the last term in your expression is $$ (x-3)^{1/5} = (-3)^{1/5} \text{.} $$ Which of the five roots of $-3$ is that? It depends on whatever conventions Mathematica has for interpreting the sequence of symbols "(-3)^(1/5)". Let's find out.

ReIm[(-3)^(1/5)]
(* {1/4 3^(1/5) (1 + Sqrt[5]), 3^(1/5) Sqrt[5/8 - Sqrt[5]/8]} *)

This is not $-(3^{1/5})$. In fact, it is the root of smallest nonnegative argument. The root your question leads me to think you were expecting is the root of third smallest nonnegative argument.

Simplify[ReIm[(-3)^(1/5) ((-1)^(2/5))^2]]
(* { -3^(1/5), 0 } *)

Why is this? Power is documented to give $x^y$ the principal value of $\mathrm{e}^{y \ln x}$. Log[-3] has the value $\mathrm{i}\pi + \ln 3$, so we are looking at $$ \mathrm{e}^{\frac{\ln 3}{5} + \mathrm{i}\frac{\pi}{5}} \text{,} $$ from which it is easy to see the argument is $\pi / 5$, so does not correspond to a real number. Consequently, $x = 0$ is not in the (real) domain of the function you give.

So to be in the (real) domain, the expression you wrote, with powers "1/5", $3x - 2 \geq 0$, $2x+1 \geq 0$, and $x-3 \geq 0$. Respectively, these given the constraints $x \geq 2/3$, $x \geq -1/2$, and $x \geq 3$. The portion of the number line satisfying all three is $x \geq 3$.

Answer 3

Or just use Reduce

Reduce[(3 x - 2)^(1/5) - (2 x + 1)^(1/5) - (x - 3)^(1/5) \[Element] 
  Reals, x, Reals]

(*   x >= 3   *)