Questions: The one-to-one function f is defined below. f(x) = 7x / (8x - 1) Find f^(-1)(x), where f^(-1) is the inverse of f. Also state the domain and range of f^(-1) in interval notation. f^(-1)(x) = Domain of f^(-1) Range of f^(-1)

The one-to-one function f is defined below. f(x) = 7x / (8x - 1) Find f^(-1)(x), where f^(-1) is the inverse of f. Also state the domain and range of f^(-1) in interval notation. f^(-1)(x) = Domain of f^(-1) Range of f^(-1)
Transcript text: The one-to-one function $f$ is defined below. \[ f(x)=\frac{7 x}{8 x-1} \] Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$. Also state the domain and range of $f^{-1}$ in interval notation. \[ f^{-1}(x)= \] Domain of $f^{-1}$ $\square$ Range of $f^{-1}$ $\square$
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Solution

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Solution Steps

Step 1: Finding the Inverse Function

To find the inverse of the function \( f(x) = \frac{7x}{8x - 1} \), we set \( y = f(x) \) and solve for \( x \) in terms of \( y \). The resulting inverse function is given by: \[ f^{-1}(x) = \frac{y}{8y - 7} \]

Step 2: Determining the Domain of \( f^{-1} \)

The domain of the inverse function \( f^{-1} \) corresponds to the range of the original function \( f(x) \). The range of \( f(x) \) is all real numbers except \( \frac{7}{8} \). Therefore, the domain of \( f^{-1} \) is: \[ \text{Domain of } f^{-1}: (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty) \]

Step 3: Determining the Range of \( f^{-1} \)

The range of the inverse function \( f^{-1} \) corresponds to the domain of the original function \( f(x) \). The domain of \( f(x) \) is all real numbers except \( \frac{1}{8} \). Therefore, the range of \( f^{-1} \) is: \[ \text{Range of } f^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty) \]

Final Answer

\[ f^{-1}(x) = \frac{y}{8y - 7} \] \[ \text{Domain of } f^{-1}: (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty) \] \[ \text{Range of } f^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty) \]

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