Answer the following questions about the function whose derivative is given belo
ID: 2884156 • Letter: A
Question
Answer the following questions about the function whose derivative is given below a. What are the critical points of f? b. On what open intervals is f increasing or decreasing? c. At what points, if any, does f assume local maximum and minimum values? f'(x) = (sin x + Squareroot 3 cos x)(sin x - cos x), 0 lessthanorequalto x lessthanorequalto 2 pi x = b. ON what open intervals is f increasing or decreasing? Select the correct choice below and fill in any answer boxes within your choice. The function f is increasing on the open interval(s) (pi/4, 2 pi/3), (5 pi/4, 5 pi/3), and decreasing on the open interval(s) (0, pi/4), (2 pi/3, 5 pi/4), (5 pi/3, 2 pi). B. The function f is decreasing on the open interval(s), and never increasing. C. The function f is increasing on the open interval(s), and never decreasing. c. At what points, if any, does f assume local maximum values? Select the correct choice below and fill in any answer boxes within your choice. The function f has a local maximum at x = 0, 2 pi/3, 5 pi/3. Three are no local maxima.Explanation / Answer
We have given f'(x)=(sinx+sqrt(3)*cosx)(sinx-cosx),for 0<=x<=2pi
a) set f'(x)=0
(sinx+sqrt(3)*cosx)(sinx-cosx)=0
sinx+sqrt(3)*cosx=0 and (sinx-cosx)=0
sinx=-sqrt(3)*cosx and sinx=cosx
tanx=-sqrt(3) and x=pi/4,5pi/4
x=2pi/3,5pi/3,pi/4,5pi/4
critical points are x=2pi/3,5pi/3,pi/4,5pi/4
b) We have f'(x)=(sinx+sqrt(3)*cosx)(sinx-cosx),for 0<=x<=2pi
at x=pi/6
f'(pi/6)=(sin(pi/6)+sqrt(3)*cos(pi/6))(sin(pi/6)-cos(pi/6))=-0.732<0
at x=pi/3
f'(pi/3)=(sin(pi/3)+sqrt(3)*cos(pi/3))(sin(pi/3)-cos(pi/3))=0.633>0
at x=3pi/4
f'(3pi/4)=(sin(3pi/4)+sqrt(3)*cos(3pi/4))(sin(3pi/4)-cos(3pi/4))=-0.732<0
at x=4pi/3
f'(4pi/3)=(sin(4pi/3)+sqrt(3)*cos(4pi/3))(sin(4pi/3)-cos(4pi/3))=0.633>0
at x=7pi/4
f'(7pi/4)=(sin(7pi/4)+sqrt(3)*cos(7pi/4))(sin(7pi/4)-cos(7pi/4))=-0.732<0
the function is increasing where f'(x)>0 and decreasing f'(x)<0 on open intervals
the function is increasing on the open intervals(pi/4,2pi/3) and (5pi/4,5pi/3), and decreasing on the open intervals (0,pi/4),(2pi/3,5pi/4),(5pi/3,2pi)
c) we have f'(x)=(sinx+sqrt(3)*cosx)(sinx-cosx),for 0<=x<=2pi
f''(x)=(sinx+sqrt(3)*cosx)*(cosx+sinx)+(sinx-cosx)*(cosx-sqrt(3)*sinx)
at x=0
f''(0)=(sqrt(3))*(1)+(-1)*(1)=sqrt(3)-1=0.732>0 concave up at x=0 function has local minimum at x=0
at x=2pi/3
f''(2pi/3)=(sin(2pi/3)+sqrt(3)*cos(2pi/3))*(cos(2pi/3)+sin(2pi/3))+(sin(2pi/3)-cos(2pi/3))*(cos(2pi/3)-sqrt(3)*sin(2pi/3))
=-2.732<0 concave down at x=2pi/3 the function has local maximum at x=2pi/3
at x=5pi/4
f''(5pi/4)=(sin(3pi/4)+sqrt(3)*cos(5pi/4))*(cos(5pi/4)+sin(5pi/4))+(sin(5pi/4)-cos(5pi/4))*(cos(5pi/4)-sqrt(3)*sin(5pi/4))
=0.732>0 concave up at x=5pi/4 the function has local minimum
at x=5pi/3
f''(5pi/3)=(sin(5pi/3)+sqrt(3)*cos(5pi/3))*(cos(5pi/3)+sin(5pi/3))+(sin(5pi/3)-cos(5pi/3))*(cos(5pi/3)-sqrt(3)*sin(5pi/3))
=-2.732<0 concave down at x=5pi/3 the function has local maximum
at x=pi/4
f''(pi/4)=(sin(pi/4)+sqrt(3)*cos(pi/4))*(cos(pi/4)+sin(pi/4))+(sin(pi/4)-cos(pi/4))*(cos(pi/4)-sqrt(3)*sin(pi/4))
=2.732>0 concave up at pi/4 the function has local minimum
the function has local maximum x=2pi/3,5pi/3 and minimum at x=pi/4,5pi/4