The absolute ma and min of x-cosx on [-pi/2, pi/2]
I got the derivative as f'(x)=1+sinx
+130466 You have the correct derivative....
To find any possible max's or min's....set the derivative to 0
1 + sinx = 0
sin x = -1 and this happens at -pi/2 on the requested interval
To see if this is a possible max or min.....take the second derivative = cosx
But putting the critical point of -pi/2 into this produces 0....which doesn't tell us anything
But this derivative is always positive for any values except at -pi/2 + n2pi [where n is an integer]....and at these points the derivative is 0........so..in simple terms.....the curve is constantly increasing from left to right except at these aforementioned points where its derivative is 0
So...this curve only has max's and min's at its enpoints....namely, a min at x= -pi/2, and a max at x = pi/2
Here's a graph of the function to show this : https://www.desmos.com/calculator/kvylqvrime
