# `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.

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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)).`