Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

/* I think the word 'Math' comes from (the) American English and the British one is 'Maths'. Please, let us know if I'm wrong about this so that, we might make the most of your overflowing knowledges respectfully... :) THANK YOU for reading. */

share|improve this question
    
I tried my best to format this, but it is very difficult to make sense of your question. –  Alex Becker Feb 29 '12 at 3:58
    
Humm... It's called "Math" and I revised the equation from the book so...?? I don't know what time is it there now but, it's already too late here (06:20)... :) OK, thank you for editing my post, now we're almost even... :D :) –  Kerim Atasoy Feb 29 '12 at 4:20
    
...I guess people're looking for more challenging problems those might worth time to spend on, maybe that's why, who knows... :) –  Kerim Atasoy Feb 29 '12 at 4:25
    
It's because "notates" and "declares" make no sense in this context. We have no idea what is being asked. –  Alex Becker Feb 29 '12 at 4:26
    
Well, I often use some dictionaries, they're quite helpful and I guess it worths spend some time with them also... :) "Don't shoot the messenger." :) –  Kerim Atasoy Feb 29 '12 at 4:33
show 4 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
:) Thank you, Gerry. See guys, I'm not that "bad" at all... :) I just need some answers, that's why. –  Kerim Atasoy Feb 29 '12 at 4:40
    
That’s pretty much the answer that I’d have given. Like you, I’m puzzled by ‘declares $[0,+\infty)$’ bit, unless it’s badly mangled from some explanation that $\lim_{x\to\infty}f(x)=0$, thereby justifying the $(0,\dots$ in the description of the range. –  Brian M. Scott Feb 29 '12 at 5:25
    
So, I can use the term "range" for that, I see now. Really, I'm not trying to make your day harder, I should be more clear as you all said before. But, actually, sometimes, I find some mistakes in the book and those might make people mad also... –  Kerim Atasoy Feb 29 '12 at 5:40
    
I can't talk or even say a word for "Limits" in Maths because I don't know that subject at all, so... –  Kerim Atasoy Feb 29 '12 at 5:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.