Domain of $\sqrt{x\sin^{2}\pi x} $

Question

$\sin^{2}$ is always positive so the only possibility we get is $x\geq0\wedge\sin^{2} (\pi x) \geq 0$ and since $\sin^{2}$ is always positive, it should be all positive real numbers then, but it is not.

Answer 1

It is true that $\sin^2(\pi x)$ is nonnegative.

But $x$ is negative if $x<0$.

That means $x\sin^2(\pi x)$ is negative if $x<0$.

So the domain is only nonnegative reals!

EDIT: I completely forgot that $\sin^2(\pi x)=0$ if $x$ is an integer!

So the domain is positive reals and negative integers.

Answer 2

if you're talking about

$$f(x)=\sqrt{x\sin^2(\pi x)}$$

case $1$: $x\geq 0$ and $\sin^2(\pi x)\geq 0$, but $\sin^2(\pi x)$ is already a nonnegative number so $x\geq 0.$

case $2$: $x\lt 0$ and $\sin^2(\pi x)=0$ but $\sin^2(\pi x)=0 \Rightarrow \sin(\pi x)=0$, which implies $\pi x=k\pi, k \in \mathbb{Z} \Rightarrow x=k \lt 0, k \in \mathbb{Z}.$ $$Dom(f)=\{x\in \mathbb{R} | x\geq 0\} \cup \{x \in \mathbb{R} | x \lt 0\ and\ x=k, k \in \mathbb{Z}\}.$$

Answer 3

The function is defined, as you correctly noted, when $x\sin^2(\pi x) \geq 0$, however this does not mean that $x\geq 0\land \sin^2(\pi x) = 0$. For example, this can also be true if $x=0$ (no matter the value of $\sin^2 (\pi x)$ and it can also be true if $\sin^2(\pi x) = 0$, no matter what the value of $x$ is.

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In general:

$$ab>0\implies (a>0\land b>0) \lor (a<0\land b<0)$$

however

$$ab\geq 0\implies (a=0)\lor(b=0)\lor(a>0\land b>0)\lor(a<0\land b<0)$$

Answer 4

We need

$$x\sin^{2}\pi x\ge0\iff x\ge0 \quad \lor \quad\sin^2 \pi x=0\iff x\ge 0 \quad \lor \quad x=k\in\mathbb{Z} $$