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

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

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Answer 1 · 2018-07-17T11:00:13

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

Explanation:

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

This is one-piece inverse with

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

Correspondingyt, the domain is given by

$x \in [\frac{1}{2} (sin (- \frac{π}{2}) - 3), \frac{1}{2} (sin (\frac{π}{2}) - 3) = [- 2, - 1]$ $x in [ 1/2 ( sin ( - pi/2 ) - 3), 1/2 ( sin ( pi/2 ) - 3 ) = [ - 2, - 1 ]$ .

See illustrative graph, within the enclosure

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

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

For information, the wholesome graph for

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

is shown below.

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

Here, g-range is without limit.

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

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

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Explanation:

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