How to find the intervals on which f is concave up? f(x)=arctan⁡(sin⁡x)f(x)=arctan⁡(sin⁡x)f′(x)=cos⁡x1+sin2xf″(x)=−(sin⁡x)(2+cos2x)(1+sin2x)2

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alkaholikd9 · Accepted Answer

For f to concave up, it&#039;s second derivative needs to be positive, so we just need to look at when f″&amp;gt;0Since (2+cos2(x)) and the denominator (1+sin2(x))2 are always positive due to the squares, what we need to look at is simply when sin⁡(x) is negative.f″(x)&amp;gt;0→sin⁡(x)&amp;lt;0→x∈((2n−1)π,(2n)π){n∈N}

Bob Huerta · Accepted Answer

A function is concave up when its second derivative is positive.  f(x)=−(sin⁡x)(2+cos2x)(1+sin2x)2  The negative sign can be taken into the numerator to give  f(x)=−−(sin⁡x)(2+cos2x)(1+sin2x)2  he denominator is always positive and the (2+cos2x) in the numerator is always positive.   This means that the sign of f′′(x) is determined by the sign of −(sin⁡x).  When sinx is negative, −(sin⁡x) is positive and hence f′′(x) is positive.  sinx is negative in the 3rd and 4th quadrants.  Hence, f′′(x) is positive when x is in the 3rd and 4th quadrants.  Hence, f(x)=arctan⁡(sin⁡x) is concave up when x is in the 3rd and 4th quadrants, such as when x∈(−5π,−4π)∪(−3π,−2π)∪(−π,0)∪(π,2π)∪(3π,4π)∪(5π,6π) and so on.

How to find the intervals on which f is concave

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