Let $\{x\}$ denote the fractional part of $x$, which is $\{x\}=x-[x]$. Let $f_{a,b}(x)=\{x+a\}+2\{x+b\}$ and let its range be $\{m_{a,b},M_{a,b})$. Find the minimum value of $M_{a,b}$ as $a$ and $b$ range along all real numbers.
I tried two methods, one is to sketch a graph which proves to be useless because there is too many things to consider if $a,b$ could be changed.
I also tried by writing $\{x\}=x-[x]$, which also seems to me to be intractable.
How should questions of this type be solved and how shall I solve this problem?
We can assume without loss of generality that $b=0$ -- adding the same constant to both $a$ and $b$ will just translate the function horizontally. Since only the fractional part of $a$ matters, we can also assume that $a\in[0,1)$.
Then we have $$f(0)=a$$ from which point it grows with a slope of $+3$ until $x=1-a$. There we have $$f(1-a-\varepsilon) = a+3(1-a-\varepsilon) = 3-2a-3\varepsilon \\ f(1-a) = 2(1-a) = 2-2a $$ and it then keeps growing with a slope of $+3$ until $$f(1-\varepsilon) = a+2-3\varepsilon$$
The highest of these two peaks is our $M_{a,0}$: $$ M_{a,0} = \max(3-2a,a+2) $$ and we can graph this as a function of $a$ to find that the minimal $M_{a,0}$ occurs when $3-2a=a+2$. This is easy to solve and gives $a=1/3$ and $M_{a,0}=2\frac13$.