How do you find the maximum value of #f(x) = sinx ( 1+ cosx) #?

1 Answer
Apr 28, 2018

Answer:

Please see the explanation below.

Explanation:

The maximum value is calculated with the first and second derivatives.

The function is

#f(x)=sinx(1+cosx)=sinx+sinxcosx #

#=sinx+1/2sin2x#

The first derivative is

#f'(x)=cosx+2xx1/2cos(2x)#

#f'(x)=0#

When

#cosx+cos2x=0#

#2cos^2x+cosx-1=0#

Solving this quadratic equation in #cosx#

The solutions are

#cosx=(-1+-sqrt((-1)^2-4(2)(-1)))/(4)=(-1+-3)/4#

#{(cosx=-1),(cosx=1/2):}#

#<=>#, #{(x=pi+2kpi),(x=pi/3+2kpi),(x=5/3pi+2kpi):}#

The second derivative is

#f''(x)=-sinx-2sin(2x)#

Therefore,

#{(f''(pi)=-0-0=0),(f''(pi/3)=-sqrt3/2-sqrt3 <0),(f''(5/3pi)=sqrt3/2+sqrt3>0):}#

When #(x=pi +2kpi)# corresponds to points of inflexions.

When #x=pi/3+2kpi# corresponds to maximum

When #x=5/3pi+2kpi# corresponds to minimum

graph{sinx+1/2sin(2x) [-2.08, 10.404, -2.845, 3.395]}