I'm teaching a precalculus course and was wanting to let my students try to solve the following problem. If
$$
f(x)=\sqrt{x}, g(x)=\frac{x}{x-1},h(x)=\sqrt[3]{x}
$$
Find the domain of
$$f\circ g\circ h
$$
We have the following.
$$
(f\circ g\circ h)(x)=f(g(\sqrt[3]{x}))=f(\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1})=\sqrt{\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}}
$$
Here is where things get interesting. To find the domain of this function, I need to find where
$$
\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\geq0
$$
And this happens when the numerator and denominator have the same sign. Solving the inequality gives me that 0 and 1 are critical points, so I have the following intervals to test for the domain
$$
(-\infty,0),(0,1),(1,\infty)
$$
Testing points in each interval shows that only the middle interval doesn't satisfy the inequality. Thus, the domain of the composition is
$$
(-\infty,0],(1,\infty)
$$
However, when I double check this with Wolframalpha and with Symbolab, they both tell me the domain is
$$
\{0\}\cup(1,\infty)
$$
I have double checked my work, and even just blindly inputting test values such as -1 gives me a valid output, so I'm wondering what is going on? On Desmos, I graphed the function and it did include the negative portion of the domain. The only thing I can think of is maybe the other sites are simplifying the expression to get
$$
\frac{x^{\frac{1}{6}}}{\sqrt{\sqrt[3]{x}-1}}
$$
and maybe they are hesitant to input negative values into the 6th root. However, if you do this, the complex numbers end up cancelling and you get a real numbered answer.
Any ideas on what is going on?
Best Answer
The fact is that Wolfram assumes the domain $x\ge 0$ for the function $\sqrt[3] x$ even if for $x \in \mathbb{R}$ the function is well defined on the whole domain.