`f(x) = sqrt(x) + cos(x/2), [0,2pi]` Use a graphing utility to graph the function and find the absolute extrema of the function on the given interval.
See the attached MATLAB and it shows that there is a maximum and a minimum within that range.
We can find the x values of these extreme points by solving the derivative of the function.
>> f = sqrt(x) + cos(x/2)
cos(x/2) + x^(1/2)
>> ezplot(f,[0, 2*pi])
1/(2*x^(1/2)) - sin(x/2)/2
It is not possible to symbolically solve this MATLAB. But we can estimate the maximu is at x = 1.7 and the minimu is at x = 5.4 radians.