How can I show the monotincity of $\frac{x}{1+|x|}$ without differential calculus?

Question

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

Answer 1

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

Answer 2

Hint:

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

Answer 3

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 $x$.

Answer 4

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

Answer 5

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

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