s mclesing and decreasing f(x)=-4cos 2x on [-.1 Select the correct choice below
ID: 2888231 • Letter: S
Question
s mclesing and decreasing f(x)=-4cos 2x on [-.1 Select the correct choice below and, if necessary, fill in the answer box(es) within your choice. A. The function is increasing on and decreasing on OB. The function is increasing on The function is never decreasing. O c. The function is decreasing on The function is never increasing. (Simplify your answers. Type your answers in interval notation. Type exact answers, using as needed. Use a comma to separate answers as needed ) (Simplify your answer. Type your answe r in interval notation. Type exact answers, using as needed. Use a comma to separate answers as needed.) (simplity your answer. Type your answer in interval notation. Type exact anawers, using x as neded. Use a comma to separate anawers as needed) to separate answers as needed.) D. The function is never increasing nor decreasing.Explanation / Answer
We have given f(x)=-4cos2(x) on [-pi,pi]
f'(x)=-4*2cos(x)*(-sin(x))
f'(x)=8cos(x)*sin(x)
plug 2cos(x)sin(x)=sin(2x)
f'(x)=4sin(2x)
to find the critical points of f(x) we need to set f'(x)=0
4sin(2x)=0
sin(2x)=0
2x=0,2x=-pi,2x=pi
critical points of f(x) are x=0,pi/2,-pi/2
f'(-2pi/3)=4sin(2*(-2pi)/3)=3.46410161514>0
f'(-pi/4)=4sin(2*(-pi/4))=-4<0
f'(pi/4)=4sin(2*(pi/4))=4>0
f'(2pi/3)=4sin(2*(2pi/3))=-3.46410161514<0
f'(x)>0 on [-pi,-pi/2) and (0,pi/2)
f(x) is increasing on interval [-pi,-pi/2] and [0,pi/2]
f'(x)<0 on (-pi/2,0) and (pi/2,pi]
f(x) is decreasing on interval (-pi/2,0) and (pi/2,pi]
so f(x) is increasing on interval [-pi,-pi/2] and [0,pi/2] and decreasing on interval (-pi/2,0) and (pi/2,pi]
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