Question

second derivative test: problem 3
(1 point)
consider the function ( f(x)=cos (x)+\frac{sqrt{2}}{2} x ). this function has two critical numbers ( a < b ) in ( 0,2 pi ). give the following:
( a= )
( b= )
( f^{prime prime}(a)= )
( f^{prime prime}(b)= )
thus ( f(x) ) has a local ( ? ) at ( a ) and a local ( ? ) at ( b ).
note: you can earn partial credit on this problem.
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Explanation:

Step1: Find the first derivative

The first derivative of \(y = f(x)=\cos(x)+\frac{\sqrt{2}}{2}x\) is \(f^{\prime}(x)=-\sin(x)+\frac{\sqrt{2}}{2}\).
Set \(f^{\prime}(x) = 0\), then \(-\sin(x)+\frac{\sqrt{2}}{2}=0\), so \(\sin(x)=\frac{\sqrt{2}}{2}\).
On the interval \([0,2\pi]\), \(x=\frac{\pi}{4}\) or \(x = \frac{3\pi}{4}\). Since \(A

Step2: Find the second derivative

The second - derivative of \(y = f(x)\) is \(f^{\prime\prime}(x)=-\cos(x)\).
When \(x = A=\frac{\pi}{4}\), \(f^{\prime\prime}(\frac{\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}<0\).
When \(x = B=\frac{3\pi}{4}\), \(f^{\prime\prime}(\frac{3\pi}{4})=-\cos(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}>0\).

Answer:

\(A=\frac{\pi}{4}\)
\(B=\frac{3\pi}{4}\)
\(f^{\prime\prime}(A)=-\frac{\sqrt{2}}{2}\)
\(f^{\prime\prime}(B)=\frac{\sqrt{2}}{2}\)
Thus \(f(x)\) has a local maximum at \(A\) and a local minimum at \(B\).