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I am not able to work out this exercise because I am not sure of the behavior of Absolute value for monotony.

Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.

Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.

Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.

Hint:

1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.


2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative

Hint:

• If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$
  

Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$

$$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$

then consider the cases

• 
  $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$
  


• 
  $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$