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baha jaff Grade: 12
        

Find the critical numbers of the function: f(x) = sin(x) / (1+cos^2(x))


on (0,2pi)      please    step by step

6 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


I will not give you the complete solution of this problem rather will give you the methodology. You try using this method and if again some problem comes pls feel free to ask the doubt. This will help in your practise.


Given:
f(x) = 2*cos(x) + sin(x)^2

f'(x) = 2*(-sin(x)) + 2*sin(x)*cos(x)
f'(x) = -2sin(x) + 2sin(x)*cos(x)
f'(x) = 2sin(x) * (cos(x) - 1)

To find the zeros:
0 = -2sin(x) + 2sin(x)*cos(x)
0 = 2* sin(x) * (-1 + cos(x))
0 = sin(x) * (cos(x) -1)

Therefore:
sin(x) = 0
x = arcsin(0) = 0 or pi or 2pi

OR

cos(x) -1 = 0
cos(x) = 1
x = arccos(1)
x = 0 or 2pi

To test concavity and look for points of inflection:
f''(x) = -2cos(x) + 2sin(x)*(-sin(x)) + cos(x)*(2cos(x))
f''(x) = -2cos(x) - 2sin(x)^2 + (2cos(x)^2)

To find the zeros:
0 = -2cos(x) - 2sin(x)^2 + (2cos(x)^2)
0 = -cos(x) - sin(x)^2 + cos(x)^2
0 = cos(x) - (1 - cos(x)^2) + cos(x)^2
0 = cos(x) - 1 + cos(x)^2 + cos(x)^2
0 = 2cos(x)^2 + cos(x) - 1
0 = (2cos(x) - 1)(cos(x) + 1)

Therefore:
0 = (2cos(x) - 1)
1 = 2cos(x)
1/2 = cos(x)
arccos(1/2) = x
x = pi/3 or 5pi/3

Therefore, the critical points are:

maximum at 0 and 2pi
minimum at pi
point of inflection at pi/3 and 5pi/3.


 








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Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi







6 years ago
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