To find the inverse of the function [tex]\( h(x) = \frac{9x}{4x - 1} \)[/tex], we need to swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and then solve for [tex]\( y \)[/tex]:
1. Replace [tex]\( h(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[
y = \frac{9x}{4x - 1}
\][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
x = \frac{9y}{4y - 1}
\][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
Multiply both sides by [tex]\( (4y - 1) \)[/tex] to clear the fraction:
[tex]\[
x(4y - 1) = 9y
\][/tex]
Distribute [tex]\( x \)[/tex] on the left side:
[tex]\[
4xy - x = 9y
\][/tex]
Gather all terms involving [tex]\( y \)[/tex] on one side of the equation:
[tex]\[
4xy - 9y = x
\][/tex]
Factor out [tex]\( y \)[/tex] from the left side:
[tex]\[
y(4x - 9) = x
\][/tex]
Divide both sides by [tex]\( (4x - 9) \)[/tex] to isolate [tex]\( y \)[/tex]:
[tex]\[
y = \frac{x}{4x - 9}
\][/tex]
Thus, the inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[
h^{-1}(x) = \frac{x}{4x - 9}
\][/tex]
Next, we need to determine the domain and range of [tex]\( h^{-1}(x) \)[/tex].
- Domain of [tex]\( h^{-1}(x) \)[/tex]: Since [tex]\( h^{-1}(x) \)[/tex] is a rational function, its domain excludes values that make the denominator zero. The denominator [tex]\( 4x - 9 \)[/tex] is zero when [tex]\( x = \frac{9}{4} \)[/tex].
Hence, the domain of [tex]\( h^{-1}(x) \)[/tex] is all real numbers except [tex]\( x = \frac{9}{4} \)[/tex]:
[tex]\[
\text{Domain of } h^{-1}(x): (-\infty, \frac{9}{4}) \cup (\frac{9}{4}, \infty)
\][/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: Because the function [tex]\( h(x) = \frac{9x}{4x - 1} \)[/tex] is a one-to-one rational function, its range is all real numbers except where the original function is undefined or unbounded. For [tex]\( h(x) \)[/tex], the point of discontinuity is where [tex]\( 4x - 1 = 0 \)[/tex], i.e., [tex]\( x = \frac{1}{4} \)[/tex]. At this point, the function tends to infinity. Therefore, the range of the original function [tex]\( h(x) \)[/tex] should correspond to the domain of the inverse function. This translates to:
[tex]\[
\text{Range of } h^{-1}(x): (-\infty, \frac{9}{4}) \cup (\frac{9}{4}, \infty)
\][/tex]
Thus, we have:
[tex]\[
h^{-1}(x) = \frac{x}{4x - 9}
\][/tex]
[tex]\[
\text{Domain of } h^{-1}(x): (-\infty, \frac{9}{4}) \cup (\frac{9}{4}, \infty)
\][/tex]
[tex]\[
\text{Range of } h^{-1}(x): (-\infty, \frac{9}{4}) \cup (\frac{9}{4}, \infty)
\][/tex]
