Confused by results of FunctionDomain

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]

• You asked for the domain where the function and x are real. The answer seems correct. If you want the domain for complex results as well, you can try FunctionDomain[(3 x - 2)^(1/5) - (2 x + 1)^(1/5) - (x - 3)^(1/5), x, Complexes] which returns True, to signify that it is defined everywhere. What was your expectation instead? Commented Dec 8, 2020 at 20:13
• @MarcoB, I believe it is defined if one chooses the real root. Have you tried using Surd? Commented Dec 8, 2020 at 20:17

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]

• Graphically, Plot[{(3 x - 2)^(1/5) - (2 x + 1)^(1/5) - (x - 3)^(1/5), Surd[3 x - 2, 5] - Surd[2 x + 1, 5] - Surd[x - 3, 5]}, {x, -5, 20}, PlotPoints -> 200, MaxRecursion -> 15, PlotStyle -> {{Thick, Red}, {Dashed, Lighter[Blue, 0.75]}}, PlotLegends -> Placed["Expressions", {.65, .7}], PlotRangePadding -> {{Automatic, Scaled[.1]}, Automatic}] Commented Dec 8, 2020 at 21:09

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 $$x \geq 3$$.

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   *)