second derivative test: problem 3\n(1 point)\nconsider the function ( f(x)=cos (x)+\frac{sqrt{2}}{2} x )…

second derivative test: problem 3\n(1 point)\nconsider 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:\n( a= )\n( b= )\n( f^{prime prime}(a)= )\n( f^{prime prime}(b)= )\nthus ( f(x) ) has a local ( ? ) at ( a ) and a local ( ? ) at ( b ).\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor

second derivative test: problem 3\n(1 point)\nconsider 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:\n( a= )\n( b= )\n( f^{prime prime}(a)= )\n( f^{prime prime}(b)= )\nthus ( f(x) ) has a local ( ? ) at ( a ) and a local ( ? ) at ( b ).\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor

Answer

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<B), (A=\frac{\pi}{4}), (B=\frac{3\pi}{4}).

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).