Question

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

Answer 1

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

Explanation:

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 [ - pi/2, pi/2 ]`.

The denominator `( 1 + x ^2 ) in [1, oo)`.

When `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.