consider the function ( f(x)=cos(x)-\frac{1}{2}x ) this function has two critical numbers ( a < b ) in (…

Explanation:

Step1: Find the first derivative

The derivative of (y = \cos(x)) is (y'=-\sin(x)) and the derivative of (y =-\frac{1}{2}x) is (y'=-\frac{1}{2}). So, (f'(x)=-\sin(x)-\frac{1}{2}). Set (f'(x) = 0), then (-\sin(x)-\frac{1}{2}=0), which gives (\sin(x)=-\frac{1}{2}). In the interval ([0,2\pi]), (x=\frac{7\pi}{6}) or (x = \frac{11\pi}{6}). So (A=\frac{7\pi}{6}), (B=\frac{11\pi}{6}).

Step2: Find the second derivative

The derivative of (f'(x)=-\sin(x)-\frac{1}{2}) is (f''(x)=-\cos(x)). Substitute (x = A=\frac{7\pi}{6}) into (f''(x)): (f''(\frac{7\pi}{6})=-\cos(\frac{7\pi}{6})=\frac{\sqrt{3}}{2}). Substitute (x = B=\frac{11\pi}{6}) into (f''(x)): (f''(\frac{11\pi}{6})=-\cos(\frac{11\pi}{6})=-\frac{\sqrt{3}}{2}).

Step3: Determine local maxima and minima

If (f''(c)>0), then (f(x)) has a local minimum at (x = c). If (f''(c)<0), then (f(x)) has a local maximum at (x = c). Since (f''(A)=\frac{\sqrt{3}}{2}>0), (f(x)) has a local minimum at (A). Since (f''(B)=-\frac{\sqrt{3}}{2}<0), (f(x)) has a local maximum at (B).

Answer:

(A=\frac{7\pi}{6}) (B=\frac{11\pi}{6}) (f''(A)=\frac{\sqrt{3}}{2}) (f''(B)=-\frac{\sqrt{3}}{2}) Local minimum at (A), local maximum at (B)