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# Find the critical numbers of the function: f(x) = sin(x) / (1+cos^2(x))on (0,2pi)      please    step by step

## 1 Answers

10 years ago

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.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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