Prove that `f(x)=sinx+sqrt(3)cosx` has maximum value at `x=pi/6` .

Explanation

We have, `f(x) = sinx + sqrt(3)cosx`
`therefore f^(')(x)=cosx + sqrt(3)(-sinx)`
`=cosx -sqrt(3)sinx`
For `f^(')(x) = 0, cosx = sqrt(3)sinx`
`rArr tanx=1/sqrt(3)=tanpi/6`
`rArr x=pi/6`
Again, differentiating `f^(')(x)`, we get
`f^(')(x) =-sinx - sqrt(3)cosx`
At x=`pi/6` `f^(')(x) = -sinpi/6-sqrt(3)cospi/6`
`=-1/2-sqrt(3).sqrt(3)/2`
`-1/2-3/2=-2 lt0`
Hence, at `x=pi/6, f(x)` has maximum value at `pi/6` is the point of local maxima.