`f(x) = sin(x) - sqrt(3)cos(x)` Consider the function on the interval (0, 2pi). Apply first derivative test to identify all relative extrema.
You need to find the relative extrema of the function, hence, you need to remember that the roots of the equation f'(x) = 0 are the extrema of the function.
You need to find the first derivative of the function:
`f'(x) = cos x + sqrt3*sin x`
You need to solve for x the equation `f'(x) = 0` :
`cos x + sqrt3*sin x = 0`
You need to divide by `cos x` , both sides:
`1 + sqrt 3*(sin x)/(cos x) = 0`
Replace `tan x` for `(sin x)/(cos x):`
`1 + sqrt 3*tan x = 0 => sqrt 3*tan x = -1 => tan x = -1/sqrt3`
You need to remember that tan x is negative for `x in (pi/2,pi)` and `x in (3pi/2,2pi).`
`x in (pi/2,pi) => x = pi - pi/6 => x = (5pi)/6 `
`x in (3pi/2,2pi) => x = 2pi - pi/6 => x = (11pi)/6`
Hence, the relative extrema of the function are the points `((5pi)/6 , f((5pi)/6 ))` and `((11pi)/6, f((11pi)/6)).`