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I have an equation of the form

$$f(x) = \sqrt{x^3 + x}$$

for which one needs to define the maximal domain, and image and domain are part of $\mathbb{R}$ (real numbers).

$$x^3 + x \geq 0 \implies x^2 \geq -1 \implies x \geq i$$

This seems a little confusing to me, since $i$ is an element of $\mathbb{C}$ and not $\mathbb{R}$.

What am I missing?

Can $x \geq i$ be a valid domain of the function given the above constraints?

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An example of a number $\geq i$ is $i$ itself. –  SDevalapurkar Apr 18 at 23:42
    
True. But what is smallest number $x > i$? –  mwhs Apr 18 at 23:43
    
$x^3+x = x(x^2+1)$, so $f$ is defined for $x\geq 0$... –  Sh3ljohn Apr 18 at 23:45
1  
You are doing two things wrong: i) You shouldn't divide by $x$ in your inequality without checking that $x\neq 0$, and for the sign of $x$; ii) Your confusion with the inequality $x\geq i$ is because this is nonsense; is the square-root function defined over negative numbers? What is the square root of $x^2$ again? –  Sh3ljohn Apr 18 at 23:59

2 Answers 2

up vote 1 down vote accepted

The domain of $\sqrt{x^3+x}$ is defined for $$x^3+x=x\times(x^2+1)$$ Since $\forall x,x^2+1>0$, we need $x>0$. I.e., the domain of $f(x)$ is simply $$\{x\in\mathbb{R}:x\geq0\}$$

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For real numbers, $x^2 + 1$ is always positive. That's not where the problem lies. The problem is the other factor in $x^3 + x = x (x^2 + 1)$.

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Thank you. But if $x^2 + 1 > 0$ (is always positive) this is equivalent to saying $x^2 > -1$, which is not part of the real numbers anymore. What do you mean with "the problem is the other factor"? –  mwhs Apr 18 at 23:51
    
@mwhs By "the problem is the other factor", he means $x$ can be negative, and that's where $f(x)$ is not defined. –  SDevalapurkar Apr 18 at 23:54
    
"this is equivalent to saying $x^2>−1$, which is not part of the real numbers anymore"; this is wrong, wrong, wrong –  Sh3ljohn Apr 18 at 23:56
    
@mwhs Really? That's the part of your sentence that you're questioning? –  Sh3ljohn Apr 19 at 0:01
1  
@mwhs $x^2>-1$ does NOT imply that $x>i$. So since $1^2>-1$, is $1>i$? Consider the unit circle in $\mathbb{C}$. –  SDevalapurkar Apr 19 at 0:05

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