To find the inverse function [tex]\( h^{-1}(x) \)[/tex] for the given one-to-one function [tex]\( h(x) = \frac{6x}{5 - 7x} \)[/tex], we will follow these steps:
1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from the function [tex]\( h(x) \)[/tex]:
[tex]\[ y = \frac{6x}{5 - 7x} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y(5 - 7x) = 6x \][/tex]
[tex]\[ 5y - 7xy = 6x \][/tex]
[tex]\[ 5y = 6x + 7xy \][/tex]
[tex]\[ 5y = x(6 + 7y) \][/tex]
[tex]\[ x = \frac{5y}{6 + 7y} \][/tex]
Thus, the inverse function [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ h^{-1}(x) = \frac{5x}{6 + 7x} \][/tex]
3. Determine the domain of [tex]\( h^{-1}(x) \)[/tex]:
- We need to find the range of the original function [tex]\( h(x) \)[/tex].
- The function [tex]\( h(x) = \frac{6x}{5 - 7x} \)[/tex] has a vertical asymptote at [tex]\( x = \frac{5}{7} \)[/tex]. For [tex]\( x \neq \frac{5}{7} \)[/tex], the function is defined for all real [tex]\( y \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(\frac{5}{7}\)[/tex], the function approaches infinity (both positive and negative).
- Specifically, the function covers all real numbers except for when the denominator equals zero.
Therefore, the range of [tex]\( h(x) \)[/tex] is all real numbers [tex]\( y \)[/tex] because the function does not have any horizontal asymptotes: [tex]\( (-\infty, \infty) \)[/tex].
Hence, the domain of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
4. Determine the range of [tex]\( h^{-1}(x) \)[/tex]:
- The range of [tex]\( h^{-1}(x) \)[/tex] should be the same as the domain of the original function [tex]\( h(x) \)[/tex].
- The domain of [tex]\( h(x) \)[/tex] is [tex]\( x \neq \frac{5}{7} \)[/tex], so [tex]\( x \)[/tex] can take any value except [tex]\( \frac{5}{7} \)[/tex].
Therefore, the range of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ (-\infty, \frac{5}{7}) \cup (\frac{5}{7}, \infty) \][/tex]
In summary:
[tex]\[ h^{-1}(x) = \frac{5x}{6 + 7x} \][/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\( (-\infty, \frac{5}{7}) \cup (\frac{5}{7}, \infty) \)[/tex]
