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I am learning about excluded values. I am faced with this problem:

$$\frac{x}{ x^2 + 9}$$

I started by trying to solve $x$ for $0$ in the denominator:

$$x^2 + 9 = 0$$

$$x^2 + 3^2 = 0$$

I then tried to break it up into a formula like $(x + a) (x - a) = 0$, however I can't seem to do this. I am starting to wonder if there are any excluded values for this problem. Any help is appreciated. Thank you for your time.

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Your assumption is correct, there are no excluded values for this problem – KBusc Aug 21 '14 at 17:21
up vote 1 down vote accepted

$x^2+9=0$ is equivalent to $x^2=-9$. Since the square of a [real] number is never negative, no such $x$ exists, so there are no excluded values.

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I assume that you are working in $\mathbb R$ (real numbers). Tell me if you are not. For any $x\in \mathbb R$ we have: $x^2\geq 0$ and consequently $x^2+9\geq 9>0$. There are no excluded values here.

Things become different when you are working in $\mathbb C$. Then $x^2+9=(x+3i)(x-3i)$ and $\pm 3i$ are values for wich the function is not defined.

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For all $x\in \mathbb{R}$: $x^2 \geq 0 \Longrightarrow x^2 + 9 \geq 9 > 0$$

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$x^2+9>0$ for all $x\in\mathbb R$. Then it's define on all $\mathbb R$.

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