Discuss the extremum of `f(x)=sinx+1/2sin2x+1/3sin3x ,0lt=xlt=pidot`

Answer

Point of global maxima :x=`pi//4`
Point of local minima : x=2pi//3`
Point of local maxima : `x=3pi//4`

Explanation

f(x) = ` sin x +1/2 sin 2x+1/3 sin 3x, 0 le x le pi`
`f(x) = cos x +cos 2x+cos3x=2 2x+cos 2x= cos 2x(2 cos x+1)`
Let f(x)=0
or cos2cx=0 or 2 cos x+1 =0
sign scheme of f(x) is as follows
Hence x =`pi//4,3pi//4` are points of maxima
`f(pi)/(4)=sin(pi)/(4)+(1)/(2)sin(pi)/(2)+(1)/(3)sin(3pi)/(4)`
`=(1)/(sqrt(2))+1/2+(1)/(sqrt(2))=(4)(3sqrt*(2))+1/2=(4sqrt(2+3)/(4))`
`f((3pi)/(4))=sin(3pi)/(4)+1/2sin((3pi)/(2))+1/3sin(9pi)/(4)`
`=(1)/(sqrt(2))-1/2+1/3(1)/(sqrt(2))=(4sqrt(2)-3)/(6)`
`f(0)=0f(pi)=0`
Thus `x=(pi)/(4)` is the point of global maxima
`x=(3pi)/(4)` is the point of local maxima and
`x=(2pi)/(3) `is the point of local minima
x=0 is he point of global maxima