How to find range of the function? f(x)=arcsin(x)/1+x^2

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Answer 1 · 2018-07-25T11:37:33

Range: [ -pi/4, pi/4 ].

arcsin x is an odd function So is y. And so, the graph is

symmetrical about O.

Graph of the given function:

graph{((1+x^2)y-arcsin x)(x^2-1)(y^2-1/16(pi)^2)=0[-3 3 -1.5 1.5]}

Note that the numerator $arcsin x \in [- \frac{π}{2}, \frac{π}{2}]$ $arcsin x in [ - pi/2, pi/2 ]$ .

The denominator $(1 + x^{2}) \in [1, \infty)$ $( 1 + x ^2 ) in [1, oo)$ .

When $y = \frac{π}{4}, arcsin x = (1 + x^{2}) \frac{π}{4}$ $y = pi/4, arcsin x = ( 1 + x^2 ) pi/4$ ,

The graph for solving this equation reveals x = 1

graph{y-arcsinx + 1/4 pi ( 1 + x^2 ) = 0}

The answer is OK.