AnnaSuxCalc
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NoMoreExams said:Well try plugging that into your calculator :)
Yay:!)
The discussion focuses on finding possible inflection points of the function f(x) = esin(x) within the interval [0, 2π]. The second derivative f''(x) is derived as f''(x) = esin(x)(cos2(x) - sin(x)). To find inflection points, the equation cos2(x) - sin(x) = 0 is solved, leading to a quadratic equation in sin(x). The solutions yield two x-values in the specified interval, specifically x = 0.6662 radians and x = 2.476 radians.
PREREQUISITESStudents and educators in calculus, mathematicians analyzing inflection points, and anyone interested in advanced trigonometric applications.
NoMoreExams said:Well try plugging that into your calculator :)
NoMoreExams said:The beauty of the anonymity of the Internetz ;-)
