What is the range of #f(x) = ln(sin^(-1)(x^2+x+3/4))# ?

1 Answer
Jun 15, 2017

Answer:

#[ln(pi/6), ln(pi/2)] ~~ [-0.6470, 0.4516]#

Explanation:

Given:

#f(x) = ln(sin^(-1)(x^2+x+3/4))#

First note that:

#x^2+x+3/4 = x^2+x+1/4+1/2 = (x+1/2)^2+1/2#

which can take any value in the range #[1/2, oo)#

The domain of #sin^(-1)# as a real valued function of real arguments is #[-1, 1]#.

So the possible valid arguments to it in #f(x)# are all in:

#[1/2, oo) nn [-1, 1] = [1/2, 1]#

#sin^(-1)(1/2) = pi/6#

#sin^(-1)(1) = pi/2#

and #sin# is monotonically increasing between these two endpoints.

So the range of #sin^(-1)(x^2+x+3/4)# is #[pi/6, pi/2]#

Hence the range of #f(x)# is:

#[ln(pi/6), ln(pi/2)]#