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Well, I've a defined function below to show and share with you guys from a textbook which is used in Math(s):

$f(x) = \frac{3}{2+x^2}$ and notates $f(-x) = f(x)$,

$f: D_f\to R_f$ and $x^2+2 \neq 0$ and $D_f=\mathbb R$ ,

for all $x \in \mathbb R: 0 < f(x) = \frac{3}{2+x^2}\leq 3 / 2, R_f=(0, 3/2]$ and declares $[0, +\infty)$

Could you help me some to analyse and understand this function equation, more particularly, for $D_f=\mathbb R$ definition. What about $x^2+2 \neq 0$ declaring, it seems like there is something missing with that $D_f$'s set assignment...? Any idea, please?

By the way, should a function ever equal a negative value for it's display ($R_f$) set?

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I can't make heads or tails of the question, but see if this helps:

We let $f(x)=3/(2+x^2)$. We note that, for all $x$, $(-x)^2=x^2$, and it follows that for all $x$, $f(-x)=f(x)$.

The denominator $x^2+2$ is never zero, so we can take the domain of the function (that's your $D_f$) to be the entire real line, $\bf R$.

For all $x$ wqe have $x^2\ge0$, so $x^2+2\ge2$, so $0\lt3/(2+x^2)\le3/2$. That is to say the range (your $R_f$) is $(0,3/2]$.

I have no idea what "declares $[0,+\infty)$" means.

Functions can definitely have negative values; this particular function doesn't.

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    I can't talk or even say a word for "Limits" in Maths because I don't know that subject at all, so...2012-02-29