I need to do this exercice without using differentiation.
$f$ is defined on $[1; +\infty[$ as $f(x)=\sqrt{x-4\sqrt{x-1}+3} + \sqrt{x-6\sqrt{x-1}+8}$
a. Show that $f$ is constant on $[5; 10]$.
b. Simplify the expression of $f(x)$ for $x\in [1; 5]$, and then for $x\in [10; +\infty]$.
Thanks!
$\sqrt{x-4\sqrt{x-1}+3}=\sqrt{x-1-4\sqrt{x-1}+4}=\sqrt{\sqrt{x-1}^2-4\sqrt{x-1}+2^2}=|\sqrt{x-1}-2|$
$\sqrt{x-6\sqrt{x-1}+8}=\sqrt{x-1-6\sqrt{x-1}+9}=\sqrt{\sqrt{x-1}^2-6\sqrt{x-1}+3^2}=|\sqrt{x-1}-3|$
$f(x)=|\sqrt{x-1}-2|+|\sqrt{x-1}-3|$
From here you have to see what happens when $x \in [1,5]$, $x \in [5,10]$, and $x \in [10,\infty)$.
For example $\sqrt{x-1}-2\ge 0$ and $\sqrt{x-1}-3 \le 0$ if $x \in [5,10]$ then $f(x)=\sqrt{x-1}-2-(\sqrt{x-1}-3)=1$