Find any relative extrema of y=arcsinx - x

Yasmin 2020-10-27 Answered
Find any relative extrema of y=arcsinxx
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comentezq
Answered 2020-10-28 Author has 106 answers
y=arcsinxx is the same as y=sin1(x)xandy=(1x2)11. When y'=0 we have a stationary point, so 1x2=1,1x2=1 and x=0, and y=0. When x is small sin1(x)=xandsin1(x)=sin1(x), so y<0 when x<0 and y>0 when x>0 and (0,0) is a point of inflection.
However, sin1(x)=sin1(180(2n1)x)orsin1((π)(2n1)x), where n is an integer, so the stationary point is cyclical. This affects f(x) but not f'(x). sin1(0)=180n where n is an integer, so f(0)=180norn(π). The stationary points are therefore "stacked" on the y axis with a separation of 180 degrees or 3.1416 radians, while x ranges from -1 to 1.
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