We need to find the points where f(x) is not differenciable. That is when the term inside the absolute value is zero. Thus, that happens at x=-1 and x=1. Since 1<=x<=4 we have that x=1 is the only non differenciable point. But is is an endpoint.
Now you need to take the derivative.
The derivative of y=|x| is y'=sgn(x) for x!=0 where sgn(x)=1 is x>0 and -1 is x<0.
Thus, the derivative of f(x) by chain rule is,
Note this can be zero when,
1)sgn(x+1) = +1 and sgn(x-1) = -1.
2)sgn(x+1) = -1 and sgn(x+1)=1.
If #1 then, sgn(x+1) = +1 for all x in [1,4] and sgn(x-1)=-1 never on [1,4].
If similary #2 does not ever equal to zero.
Thus, there are no critical points inside the open interval.
Just check the endpoints.