What is the Range of the Function f(x) = (1+x)^0.6 / (1+x^0.6) for x in [0,1]?

[juantheron]

range of function $\displaystyle f(x) = \frac{(1+x)^{0.6}}{1+x^{0.6}}\;\forall x \in \left[0,1\right]$

 

[Fernando Revilla]

My solution:

Spoiler

(a) $f(0)=1,\quad f(1)=\dfrac{1}{\sqrt[5]{4}}<1.$
(b) $f'(x)=\ldots=0.6\cdot\dfrac{x^{0.4}-1}{x^{0.4}(x+1)^{0.4}(x^{0.6}+1)^2}<0\quad \forall x\in[0,1]\Rightarrow f$ is strictly deccreasing on $[0,1].$
(c) By the Intermediate Value Theorem, $$\boxed{\;\text{range }f=\left[\dfrac{1}{\sqrt[5]{4}},1\right]\;}$$

 

[juantheron]

Thanks Feranado Revilla for nice solution

My Solution.

Spoiler

Using $(1+x)^{p}\leq 1+x^{p}\;,$ where $0<p<1$

So $(1+x)^{0.6}\leq 1+x^{0.6}$ and equality hold when $x=0$

So we get $\displaystyle \frac{(1+x)^{0.6}}{1+x^{0.6}}\leq 1$ at $x=0$

And Using power mean inequality $\displaystyle \left(\frac{1+x}{2}\right)^{0.6} \geq \frac{1+x^{0.6}}{2}\Rightarrow \frac{(1+x)^{0.6}}{1+x^{0.6}}\geq 2^{-0.4}$ at $x=1$

So we get $\displaystyle \frac{(1+x)^{0.6}}{1+x^{0.6}}\in \left[2^{-0.4},1\right]$