# What is the domain of g(x) = sin^-1(2x + 3)?

Jul 17, 2018

$x \in \left[- 2 , - 1\right]$ and $y \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, for this truncated graph.

#### Explanation:

$g \left(x\right) = {\sin}^{- 1} \left(2 x + 3\right)$.

This is one-piece inverse with

$g \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ of $x = \frac{1}{2} \left(\sin g - 3\right)$.

Correspondingyt, the domain is given by

x in [ 1/2 ( sin ( - pi/2 ) - 3), 1/2 ( sin ( pi/2 ) - 3 ) = [ - 2, - 1 ].

See illustrative graph, within the enclosure

$x = - 2 , y = \frac{\pi}{2} , x = - 1 \mathmr{and} y = - \frac{\pi}{2}$o.

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

For information, the wholesome graph for

$g = {\left(\sin\right)}^{- 1} \left(2 x + 3\right)$, using the inverse $x = \frac{1}{2} \sin \left(\left(g\right) - 3\right)$

is shown below.

graph{x-1/2( sin (y) - 3 ) = 0 [-3 0 -10 10]}

Here, g-range is without limit.

I use ${\left(\sin\right)}^{- 1}$ for the wholesome inverse. This enables me to

reveal more details..