Find the absolute maximum and absolute minimum of g(x)=sin(x)-x on [-pi/2 , pi/2]

g(x)=sin(x) - x        take the derivative and set it to 0

g' (x)  = cosx - 1  = 0      add 1 to each side

cosx = 1      and this happens at 0  on [-pi/2, pi/2]

And the slope of this curve is always negative except at  n*2pi   where n is an integer.......at these points, the slope is always  0

So.....this has no absolute max's or min's  on the interval [-pi/2 , pi/2]

Here's the graph that confirms this :  https://www.desmos.com/calculator/6pcoba738r