Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$ 
Find minimum and maximum of $$A = \sqrt{x-2} +  2\sqrt{x+1} + 2019 -x$$

How to solve this problem without using derivatives?

Michael Rozenberg's edit: 
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$
$$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$
$$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$
which gives $x_{max}=3.$

Let $f(x) = \sqrt{x-2} +  2\sqrt{x+1} + 2019 -x$ then
$$
f'(x) = \frac{1}{2\sqrt{x - 2}} + \frac{1}{\sqrt{x+1}} - 1
$$
Equating this to zero for real $x$, we get $x = 3$. The second derivative is negative hence the global maximum is $f(3) = 2021.$
