Answer :
To find the inverse function [tex]\( g^{-1}(x) \)[/tex] of the given one-to-one function [tex]\( g(x)=\frac{-2x + 3}{1 + 3x} \)[/tex], and to determine its domain and range, we can follow these steps:
1. Set up the equation for the inverse:
[tex]\[ y = \frac{-2x + 3}{1 + 3x} \][/tex]
We need to solve this equation for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to represent [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ x = \frac{-2y + 3}{1 + 3y} \][/tex]
3. Solve for [tex]\( y \)[/tex]:
- First, clear the denominator by multiplying both sides by [tex]\( 1 + 3y \)[/tex]:
[tex]\[ x(1 + 3y) = -2y + 3 \][/tex]
- Expand and rearrange the equation:
[tex]\[ x + 3xy = -2y + 3 \][/tex]
- Group all [tex]\( y \)[/tex]-terms on one side of the equation:
[tex]\[ 3xy + 2y = 3 - x \][/tex]
- Factor out [tex]\( y \)[/tex] on the left side:
[tex]\[ y(3x + 2) = 3 - x \][/tex]
- Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3 - x}{3x + 2} \][/tex]
Therefore, the inverse function [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ g^{-1}(x) = \frac{3 - x}{3x + 2} \][/tex]
4. Determine the domain of [tex]\( g^{-1}(x) \)[/tex]:
The domain of [tex]\( g^{-1}(x) \)[/tex] is the range of [tex]\( g(x) \)[/tex]. We observe the behavior of the original function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{-2x + 3}{1 + 3x} \][/tex]
- The denominator [tex]\( 1 + 3x \)[/tex] must not be zero, so [tex]\( 3x \neq -1 \)[/tex] which implies [tex]\( x \neq -\frac{1}{3} \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( g(x) \to -\frac{2}{3} \)[/tex].
- As [tex]\( x \to -\frac{1}{3}^+ \)[/tex], [tex]\( g(x) \to +\infty \)[/tex].
- As [tex]\( x \to -\frac{1}{3}^- \)[/tex], [tex]\( g(x) \to -\infty \)[/tex].
Thus, the range of [tex]\( g(x) \)[/tex] excludes [tex]\( -\frac{2}{3} \)[/tex]. Practically, as [tex]\( g(x) \to \infty \text{ and } g(x) \to -\infty \)[/tex], it will cover all real values except [tex]\( -\frac{2}{3} \)[/tex] where the function is undefined.
Conclusively, the range of [tex]\( g(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) = \mathbb{R} \][/tex]
Since values near [tex]\(-\frac{2}{3}\)[/tex] are effectively covered by extremities, the domain of [tex]\( g^{-1}(x) \)[/tex] is from [tex]\(-\infty\)[/tex] to [tex]\(\infty\)[/tex], excluding where its denominator is zero, thus:
The effective domain:
[tex]\[ (-\infty, -\frac{1}{3}) \cup (-\frac{1}{3}, \infty) = (-\frac{1}{3}, \infty) \][/tex]
5. Determine the range of [tex]\( g^{-1}(x) \)[/tex]:
The range of [tex]\( g^{-1}(x) \)[/tex] is the domain of [tex]\( g(x) \)[/tex], which is all real numbers except [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ (-\infty, 3] \cup (3, \infty) \][/tex]
Practically, the form places it:
[tex]\(\mathbb{R}\)[/tex] effectively nuanced between general divisions.
Therefore, the inverse [tex]\( g^{-1}(x) \)[/tex], its domain, and its range are as follows:
[tex]\[ g^{-1}(x) = \frac{3 - x}{3x + 2} \][/tex]
[tex]\[ \text{Domain}: (-\frac{1}{3}, \infty) \][/tex]
\[
\text{Range}: (-\infty, 1.5)
1. Set up the equation for the inverse:
[tex]\[ y = \frac{-2x + 3}{1 + 3x} \][/tex]
We need to solve this equation for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to represent [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ x = \frac{-2y + 3}{1 + 3y} \][/tex]
3. Solve for [tex]\( y \)[/tex]:
- First, clear the denominator by multiplying both sides by [tex]\( 1 + 3y \)[/tex]:
[tex]\[ x(1 + 3y) = -2y + 3 \][/tex]
- Expand and rearrange the equation:
[tex]\[ x + 3xy = -2y + 3 \][/tex]
- Group all [tex]\( y \)[/tex]-terms on one side of the equation:
[tex]\[ 3xy + 2y = 3 - x \][/tex]
- Factor out [tex]\( y \)[/tex] on the left side:
[tex]\[ y(3x + 2) = 3 - x \][/tex]
- Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3 - x}{3x + 2} \][/tex]
Therefore, the inverse function [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ g^{-1}(x) = \frac{3 - x}{3x + 2} \][/tex]
4. Determine the domain of [tex]\( g^{-1}(x) \)[/tex]:
The domain of [tex]\( g^{-1}(x) \)[/tex] is the range of [tex]\( g(x) \)[/tex]. We observe the behavior of the original function [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{-2x + 3}{1 + 3x} \][/tex]
- The denominator [tex]\( 1 + 3x \)[/tex] must not be zero, so [tex]\( 3x \neq -1 \)[/tex] which implies [tex]\( x \neq -\frac{1}{3} \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( g(x) \to -\frac{2}{3} \)[/tex].
- As [tex]\( x \to -\frac{1}{3}^+ \)[/tex], [tex]\( g(x) \to +\infty \)[/tex].
- As [tex]\( x \to -\frac{1}{3}^- \)[/tex], [tex]\( g(x) \to -\infty \)[/tex].
Thus, the range of [tex]\( g(x) \)[/tex] excludes [tex]\( -\frac{2}{3} \)[/tex]. Practically, as [tex]\( g(x) \to \infty \text{ and } g(x) \to -\infty \)[/tex], it will cover all real values except [tex]\( -\frac{2}{3} \)[/tex] where the function is undefined.
Conclusively, the range of [tex]\( g(x) \)[/tex]:
[tex]\[ (-\infty, -\frac{2}{3}) \cup (-\frac{2}{3}, \infty) = \mathbb{R} \][/tex]
Since values near [tex]\(-\frac{2}{3}\)[/tex] are effectively covered by extremities, the domain of [tex]\( g^{-1}(x) \)[/tex] is from [tex]\(-\infty\)[/tex] to [tex]\(\infty\)[/tex], excluding where its denominator is zero, thus:
The effective domain:
[tex]\[ (-\infty, -\frac{1}{3}) \cup (-\frac{1}{3}, \infty) = (-\frac{1}{3}, \infty) \][/tex]
5. Determine the range of [tex]\( g^{-1}(x) \)[/tex]:
The range of [tex]\( g^{-1}(x) \)[/tex] is the domain of [tex]\( g(x) \)[/tex], which is all real numbers except [tex]\( x = -\frac{1}{3} \)[/tex]:
[tex]\[ (-\infty, 3] \cup (3, \infty) \][/tex]
Practically, the form places it:
[tex]\(\mathbb{R}\)[/tex] effectively nuanced between general divisions.
Therefore, the inverse [tex]\( g^{-1}(x) \)[/tex], its domain, and its range are as follows:
[tex]\[ g^{-1}(x) = \frac{3 - x}{3x + 2} \][/tex]
[tex]\[ \text{Domain}: (-\frac{1}{3}, \infty) \][/tex]
\[
\text{Range}: (-\infty, 1.5)
