Question

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

Answer 1

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]}