Discuss the extremum of `f(x)=1+2sinx+3` `cos^2x ,xlt=xlt=(2pi)/3`

Answer

Point of global minima:`x=pi//3`
Point of global maxima : x= `sin^(-1)(1//3)`

Explanation

`f(x) =1+2sinx+3 cos^(2)x,0 le x le 2 pi//3`
Therefore f(x) = 2 cosx-6 sinx cos x
`=2cos x (1-3 sinx)`
`f'(x) =0 rarrr cos x =0 or 1-3 sin x =0`
i.e `x=pi//2 or sin x =1//3`
Hence x =`pi//2` is is point of minima
`f''sin^(-1) 1//3` is pint of maximum
`f_(min)=f(pi)/(2)=1+2 sin (pi)/(2) +3 cos^(2)(pi)/(2)=1+2=3`
`f_(max)=f(sin^(-1)(1)/(3))=1+2(1/3)+3(1-(1)/(9))=5/3+8/3=13/3`