Consider the function on the interval (0, 2π). f (x)= sin (x) + cos (x) Apply the First Derivative Test to identify the relative extrema. relative maximum (x, y) = ( ) relative minimum (x, y) = ( )

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You need to evaluate the derivative of the given function such that:

`f'(x) = (sin x + cos x)'`

`f'(x) = cos x - sin x`

You need to solve the equation `f'(x) = 0`  such that:

`cos x - sin x = 0`

You should divide by `cos x`  such that:

`1 - sin x/cos x = 0 => -sin x/cos x = -1 => sin x/cos x = 1`

You need to remember that `sin x/cos x = tan x` , hence, substituting `tan x`  for `sin x/cos x`  yields:

`tan x = 1`

The tangent is positive in quadrants 1 and 3 for `x in (0,2pi` ) such that:

`x = pi/4`  quadrant 1

`x = pi+pi/4 = 5pi/4 ` quadrant 3

Notice that the derivative of the function is negative if `x in(pi/4, pi/4)`  and positive if `x in (0,pi/4)U(5pi/4, 2pi).`

You need to evaluate the values of function at critical values `x=pi/4`  and `x=5pi/4 ` such that:

`f(pi/4) = sin (pi/4) + cos(pi/4) = sqrt2/2 + sqrt2/2 = sqrt2`

`f(5pi/4) = sin(5pi/4) + cos(5pi/4) = -sqrt2`

Hence, evaluating the relative extrema of the function yields that it has a relative maximum at `(pi/4, sqrt2)`  and it has a relative minimum at `(5pi/4,-sqrt2).`

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