Find any relative extrema of y=arcsinx - x

Question

Find the derivative and then set it to zero and find the relative extrema

Answer 1

y=arcsinx-x is the same as y= sin^-1(x)-x and y'=(sqrt(1-x^2))^-1-1. When y'=0 we have a stationary point, so sqrt(1-x^2)=1, 1-x^2=1 and x=0, and y=0. When x is small sin^-1(x)=x and sin^-1(x)=-sin^-1(-x), so y<0 when x<0 and y>0 when x>0 and (0,0) is a point of inflection.

However, sin^-1(x)=sin^-1(180(2n-1)-x) or sin^-1((pi)(2n-1)-x), where n is an integer, so the stationary point is cyclical. This affects f(x) but not f'(x). sin^-1(0)=180n where n is an integer, so f(0)=180n or n(pi). 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.