The range of the function $f(x) = \frac{(1+x)^4}{1+x^4}$ where $f: \mathbb{R} \to \mathbb{R}$ is _____
Based on desmos.com the range is $[0,8]$
On solving we get $f'(x) = \frac{-4(x+1)(x-1)(x+1)^2(1+x+x^2)}{(1+x^4)^2}$
We get local minimum at $x=-1$ and local maximumn at $x=1$
Which helps us to get the required range but how we will prove that it is bounded function
$f(x) = \frac{(1+x)^4}{1+x^4}$
$f'(x) = \frac{-4(x+1)(x-1)(x+1)^2(1+x+x^2)}{(1+x^4)^2}$
As, $\frac{(x+1)^2(1+x+x^2)}{(1+x^4)^2}\ge0$ for any value of $x$.
So, $f'(x)<0$ i.e. $f(x)$ is decreasing for $x>1$
$f'(x)>0$ i.e. $f(x)$ is increasing for $x<1$
$\lim_{x\to \pm\infty}\frac{(1+x)^4}{1+x^4}=\lim_{x\to \pm\infty}\frac{(\frac{1}x+1)^4}{1+(\frac{1}x)^4}=1$
So, $f(x)$ is a bounded function.
Hence, one end point of range $f(1)=8$ because function started decreasing from $x=1$ and stoped at $f(x=\infty)=1$. Similarly, another end point of range $f(-1)=0$ because function started increasing from $x=1$ and stoped at $f(x=-\infty)=1$ Therefore, Range of the function is $[f(-1),f(1)]=[0,8]$
