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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```
7 years ago

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.

All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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