How find the range of the $t_{1}-t_{2}$? let
$$f(x)=\begin{cases}
\dfrac{1}{a-1}(x-1)&x\ge a\\
\dfrac{1}{a-2}(x-2)& x<a
\end{cases}$$
There exist $t_{1},t_{2}$ such that 
$$f(t_{1})=\dfrac{1}{2},f(t_{2})=\dfrac{3}{2}$$
then $$t_{1}-t_{2}$$ the range of values is ?
my book have only have  result $$t_{1}-t_{2}\in (-\infty,-\dfrac{1}{2})\cup (\dfrac{1}{2},+\infty)$$  true or false?

if $a \geq 2$, then : 
since $f(x)$ is greater than one only for $x \geq a$ and is less that one only for $x < a$. $f(t_1) = \frac12$ only for $x < a$ and $f(t_2) = \frac32$ only for $x \geq a$. 
 so $ \frac{t_1-2}{a-2} = \frac12 $ and $ \frac{t_2-1}{a-1} = \frac32 $, then $t_1-t_2 = -a + \frac32$ and the rest is left to you. Similarly you can do it for other values of $a$.
