To find the inverse of the one-to-one function [tex]\( g(x) = \frac{7x}{8x - 5} \)[/tex] and determine the domain and range of [tex]\( g^{-1}(x) \)[/tex], we will follow these steps:
### Step 1: Set up the equation to find the inverse
We begin by setting [tex]\( g(x) = y \)[/tex] and solving for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{7x}{8x - 5} \][/tex]
### Step 2: Solve for [tex]\( x \)[/tex]
To solve for [tex]\( x \)[/tex], we will cross-multiply and isolate [tex]\( x \)[/tex]:
[tex]\[ y(8x - 5) = 7x \][/tex]
[tex]\[ 8yx - 5y = 7x \][/tex]
[tex]\[ 8yx - 7x = 5y \][/tex]
[tex]\[ x(8y - 7) = 5y \][/tex]
[tex]\[ x = \frac{5y}{8y - 7} \][/tex]
So, the inverse function [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ g^{-1}(x) = \frac{5x}{8x - 7} \][/tex]
### Step 3: Determine the domain of [tex]\( g^{-1}(x) \)[/tex]
The domain of [tex]\( g^{-1}(x) \)[/tex] is all values of [tex]\( x \)[/tex] for which the expression is defined. The denominator [tex]\( 8x - 7 \)[/tex] cannot be zero:
[tex]\[ 8x - 7 \ne 0 \][/tex]
[tex]\[ 8x \ne 7 \][/tex]
[tex]\[ x \ne \frac{7}{8} \][/tex]
Therefore, the domain of [tex]\( g^{-1}(x) \)[/tex] is all real numbers except [tex]\( \frac{7}{8} \)[/tex]:
[tex]\[ \text{Domain of } g^{-1}(x) : (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty) \][/tex]
### Step 4: Determine the range of [tex]\( g^{-1}(x) \)[/tex]
The range of [tex]\( g^{-1}(x) \)[/tex] corresponds to the domain of the original function [tex]\( g(x) \)[/tex]. From the original function [tex]\( g(x) = \frac{7x}{8x - 5} \)[/tex], we know the function is not defined when [tex]\( 8x - 5 = 0 \)[/tex]:
[tex]\[ 8x - 5 \ne 0 \][/tex]
[tex]\[ 8x \ne 5 \][/tex]
[tex]\[ x \ne \frac{5}{8} \][/tex]
Thus, [tex]\( g(x) \)[/tex] is defined for all real numbers except [tex]\( \frac{5}{8} \)[/tex]. The range of [tex]\( g(x) \)[/tex] is all real numbers except the value [tex]\( 0 \)[/tex] since [tex]\( g(x) \)[/tex] cannot be zero when [tex]\( x = \frac{0}{7} = 0 \)[/tex]:
[tex]\[ \text{Range of } g(x) : (-\infty, 0) \cup (0, \infty) \][/tex]
Therefore, the range of [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ \text{Range of } g^{-1}(x) : (-\infty, 0) \cup (0, \infty) \][/tex]
### Conclusion
The inverse function [tex]\( g^{-1}(x) \)[/tex], the domain, and the range of [tex]\( g^{-1}(x) \)[/tex] are as follows:
[tex]\[ g^{-1}(x) = \frac{5x}{8x - 7} \][/tex]
[tex]\[ \text{Domain of } g^{-1}(x) : (-\infty, 0) \cup (0, \infty) \][/tex]
[tex]\[ \text{Range of } g^{-1}(x) : (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty) \][/tex]
