`f(x) = sin^2(x) + sin(x)` Consider the function on the interval (0, 2pi). Apply first derivative test to identify all relative extrema.
You need to evaluate the relative extrema of the function, hence, you need to find the solutions to the equation `f'(x) = 0` .
You need to determine the first derivative, using the chain rule, such that:
`f'(x) = (sin^2 x + sin x)' => f'(x) = 2sin x*cos x + cos x`
You need to solve for x the equation f'(x) = 0:
`2sin x*cos x + cos x = 0`
Factoring out cos x yields:
`cos x*(2sin x + 1) = 0`
`cos x = 0`
You need to remember that the cosine function is 0, in interval `(0,2pi), ` for `x = pi/2` and `x = 3pi/2` .
`2 sin x + 1 = 0 => 2sin x = -1 => sin x = -1/2`
You need to remember that sine function is negative on intervals `(pi,3pi,2)` and `(3pi/2,2pi).`
`x = pi + pi/6 => x = (7pi)/6`
`x = 2pi - pi/6 => x = (11pi)/6`
Hence, evaluating the relative extrema of the function yields `((7pi)/6, f((7pi)/6))` and `((11pi)/6,f((11pi)/6)).`