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

Jun 15, 2017

$\left[\ln \left(\frac{\pi}{6}\right) , \ln \left(\frac{\pi}{2}\right)\right] \approx \left[- 0.6470 , 0.4516\right]$

#### Explanation:

Given:

$f \left(x\right) = \ln \left({\sin}^{- 1} \left({x}^{2} + x + \frac{3}{4}\right)\right)$

First note that:

${x}^{2} + x + \frac{3}{4} = {x}^{2} + x + \frac{1}{4} + \frac{1}{2} = {\left(x + \frac{1}{2}\right)}^{2} + \frac{1}{2}$

which can take any value in the range $\left[\frac{1}{2} , \infty\right)$

The domain of ${\sin}^{- 1}$ as a real valued function of real arguments is $\left[- 1 , 1\right]$.

So the possible valid arguments to it in $f \left(x\right)$ are all in:

$\left[\frac{1}{2} , \infty\right) \cap \left[- 1 , 1\right] = \left[\frac{1}{2} , 1\right]$

${\sin}^{- 1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$

${\sin}^{- 1} \left(1\right) = \frac{\pi}{2}$

and $\sin$ is monotonically increasing between these two endpoints.

So the range of ${\sin}^{- 1} \left({x}^{2} + x + \frac{3}{4}\right)$ is $\left[\frac{\pi}{6} , \frac{\pi}{2}\right]$

Hence the range of $f \left(x\right)$ is:

$\left[\ln \left(\frac{\pi}{6}\right) , \ln \left(\frac{\pi}{2}\right)\right]$