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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = sinx)^2 - sqrt(3)*sinx     0<x<pi

To find critical values, we wil need to determine first derivative.

f'(x) = 2sinx*cosx - sqrt(3)*cosx

Let us factor:

f'(x)=cosx ( 2sinx - sqrt3)

Critical values are the derivative's zeros.

==> cosx (2sinx- sqrt3) = 0

==> cosx = 0 ==> x= pi/2

==> 2sinx-sqrt3 = 0 ==> sinx= sqrt3/2 ==> x2= pi/3, 5pi/6

Then critical values are: x= ( pi/2, pi/3, 2pi/3}

Then extreme values are:

f(pi/2) = (sinpi/2)^2 - sqrt3*sinpi/2 = 1- sqrt3

f(pi/3)= (sin(pi/3)^2 - sqrt3*sin(pi/3) = 3/4 -  3/2= -3/4

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the critical points of f(x) = (sinx)^2-sqrt(3)sinx , (0 < x< pi.

We differentiate f(x) and equate f'(x) = 0. Then we solve for x in f'(x) = 0. The solutions so obtained are the critical points of x.

f(x) = sin ^2x - sqrt(3)*sinx.

Differntiating both sides, we get:

f'(x) = (sin^2x - sqrt(3)*sinx)'

f'(x) = (sin^2x)- sqrt3* (sinx)'

f(x) = 2sinx*cosx - sqrt(3) cosx.

Therefore f'(x) = 0 gives:

2sinxcosx-sqrt(3)cosx = 0

(2sinx-sqrt3)(cosx) = 0

Therefore 2sinx -sqrt3 = 0 or cosx = 0.

2sinx-sqrt3 = 0 gives: sinx = sqrt3/2.

Therefore x = pi/3 or x = 2pi/3.

cosx = 0 gives: x = pi/2.

So the solution set of critical points x is {pi/3 , pi/2 , 2pi/3} for 0 < x <  pi.

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