Question

consider the function ( f(x)=cos(x)-\frac{1}{2}x ) this function has two critical numbers ( a < b ) in ( 0,2pi ). give the following:

( a=)

( b=)

( f(a)=)

( f(b)=)

thus ( f(x) ) has a local at ( a ) and a local at ( b ).

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 \(B\)