Determine the images of the functions f: R -> R....?

[rayne1]

Determine the images of functions f: R -> R defined as follows:
a) f(x) = x^2/(1+x^2)
b) f(x) = x/(1+|x|)

I have no idea if I am doing it right but this is what I did for a):

The two sets:
f: A -> B
The image of f is the set b ∈ B such that f(x) = b has a solution.
Since f(x) = x^2/(1+x^2) and f(x) = b, we have b = x^2/(1+x^2).
Then, b(1+x^2) = x^2.
For this to be true b >= 0. So, does that mean the image of f is b>=0? If so, is there anything more I need to show?

 

[Evgeny.Makarov]

Since f(x) = x^2/(1+x^2) and f(x) = b, we have b = x^2/(1+x^2).
Then, b(1+x^2) = x^2.
For this to be true b >= 0. So, does that mean the image of f is b>=0?

You are right that $b=x^2/(1+x^2)$ implies $b\ge0$, but the converse implication does not always hold. In $x^2/(1+x^2)$, the numerator is smaller than the denominator, so $b<1$. Consider small $x$ and large $x$. Can $b$ assume any value between $0$ and $1$?

For $f(x)=x/(1+|x|)$, one can see similarly that $|x|<1+|x|$, so $|f(x)|=|x|/(1+|x|)<1$, which means that $-1<f(x)<1$. Note that $f(x)$ is odd, i.e., $f(x)=-f(-x)$, so it's enough to study its behavior when $x>0$. Can $f(x)$ assume any value between 0 and 1 (and therefore between $-1$ and $1$?

 

[rayne1]

You are right that $b=x^2/(1+x^2)$ implies $b\ge0$, but the converse implication does not always hold. In $x^2/(1+x^2)$, the numerator is smaller than the denominator, so $b<1$. Consider small $x$ and large $x$. Can $b$ assume any value between $0$ and $1$?

For $f(x)=x/(1+|x|)$, one can see similarly that $|x|<1+|x|$, so $|f(x)|=|x|/(1+|x|)<1$, which means that $-1<f(x)<1$. Note that $f(x)$ is odd, i.e., $f(x)=-f(-x)$, so it's enough to study its behavior when $x>0$. Can $f(x)$ assume any value between 0 and 1 (and therefore between $-1$ and $1$?

So for a) it's 0 <= b < 1 and for b) the image can just be written as -1 < f(x) < 1?

 

[Evgeny.Makarov]

So for a) it's 0 <= b < 1 and for b) the image can just be written as -1 < f(x) < 1?

Yes. I would use $f(x)$ instead of $b$ for a) as well.