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)=7x8x−1 f(x) = \frac{7x}{8x - 1} , we set y=f(x) y = f(x) and solve for x x in terms of y y . The resulting inverse function is given by: f−1(x)=y8y−7 f^{-1}(x) = \frac{y}{8y - 7}

Step 2: Determining the Domain of f−1 f^{-1}

The domain of the inverse function f−1 f^{-1} corresponds to the range of the original function f(x) f(x) . The range of f(x) f(x) is all real numbers except 78 \frac{7}{8} . Therefore, the domain of f−1 f^{-1} is: Domain of f−1:(−∞,78)∪(78,∞) \text{Domain of } f^{-1}: (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty)

Step 3: Determining the Range of f−1 f^{-1}

The range of the inverse function f−1 f^{-1} corresponds to the domain of the original function f(x) f(x) . The domain of f(x) f(x) is all real numbers except 18 \frac{1}{8} . Therefore, the range of f−1 f^{-1} is: Range of f−1:(−∞,18)∪(18,∞) \text{Range of } f^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty)

Final Answer

f−1(x)=y8y−7 f^{-1}(x) = \frac{y}{8y - 7} Domain of f−1:(−∞,78)∪(78,∞) \text{Domain of } f^{-1}: (-\infty, \frac{7}{8}) \cup (\frac{7}{8}, \infty) Range of f−1:(−∞,18)∪(18,∞) \text{Range of } f^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty)

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