# Critical points / extreme valuesDetermine the critical points and the local extreme values. f(x)=(sinx)^2-[sqrt(3)*sinx] 0 < x < pi

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### 1 Answer

To determine the local extreme values of the function, we'll have to do the first derivative test.

For this reason, we'll determine the expression of the first derivative:

f'(x) = {(sinx)^2-[sqrt(3)*sinx]}'

f'(x) = 2sin x*cos x - sqrt3*cos x

Now, to calculate the local extreme of the function, we'll determine the roots of the first derivative:

f'(x) = 0

2sin x*cos x - sqrt3*cos x = 0

We'll factorize by cos x:

cos x(2sin x - sqrt 3) = 0

We'll put each factor as zero:

cos x = 0

This in an elementary equation:

x = arccos 0

x = pi/2

2sin x - sqrt 3 = 0

2sin x = sqrt3

sin x = sqrt 3/2

x = arcsin (sqrt 3/2)

x = pi/3

x = pi - pi/3

x = 2pi/3

Critical values of x: {pi/3 , pi/2 , 2pi/3}.

The local extremes of the function are the points whose x coordinate has the values:{pi/3 , pi/2 , 2pi/3}.

Now, we'll determine the local extremes. For this reason, we'll substitute x by the critical values:

f(pi/3) = (sin pi/3)^2 - sqrt3*sin pi/3

f(pi/3) = 3/4 - 3/2

f(pi/3) =-3/4

f(pi/2) = (sin pi/2)^2 - sqrt3*sin pi/2

f(pi/2) = 1 - sqrt 3

f(2pi/3) = (sin 2pi/3)^2 - sqrt3*sin 2pi/3

f(2pi/3) = f(pi/3) = -3/4