Questions: The one-to-one function h is defined below. h(x) = (7x-1)/(3x+7) Find h^(-1)(x), where h^(-1) is the inverse of h. Also state the domain and range of h^(-1) in interval notation. h^(-1)(x) = (3x)/(x-7) Domain of h^(-1): (-∞, 7) ∪ (7, ∞) Range of h^(-1):

The one-to-one function h is defined below.

h(x) = (7x-1)/(3x+7)

Find h^(-1)(x), where h^(-1) is the inverse of h.
Also state the domain and range of h^(-1) in interval notation.

h^(-1)(x) = (3x)/(x-7)
Domain of h^(-1): (-∞, 7) ∪ (7, ∞)
Range of h^(-1):

Transcript text: The one-to-one function $h$ is defined below. \[ h(x)=\frac{7 x-1}{3 x+7} \] Find $h^{-1}(x)$, where $h^{-1}$ is the inverse of $h$. Also state the domain and range of $h^{-1}$ in interval notation. \[ \begin{array}{l} \qquad h^{-1}(x)=\frac{3 x}{x-7} \\ \text { Domain of } h^{-1}:(-\infty, 7) \cup(7, \infty) \\ \text { Range of } h^{-1}: \square \end{array} \]

Solution

Solution Steps

To find the inverse of the function $ h(x) = \frac{7x - 1}{3x + 7} $, we need to follow these steps:

Replace $ h(x) $ with $ y $: $ y = \frac{7x - 1}{3x + 7} $.
Swap $ x $ and $ y $ to get $ x = \frac{7y - 1}{3y + 7} $.
Solve for $ y $ in terms of $ x $.
The resulting expression for $ y $ will be $ h^{-1}(x) $.
Determine the domain and range of $ h^{-1}(x) $.

Solution Approach

Replace $ h(x) $ with $ y $: $ y = \frac{7x - 1}{3x + 7} $.
Swap $ x $ and $ y $: $ x = \frac{7y - 1}{3y + 7} $.
Solve for $ y $ in terms of $ x $.
The resulting expression for $ y $ will be $ h^{-1}(x) $.
Determine the domain and range of $ h^{-1}(x) $.

Step 1: Find the Inverse Function

To find the inverse of the function $ h(x) = \frac{7x - 1}{3x + 7} $, we start by setting $ y = h(x) $: \[ y = \frac{7x - 1}{3x + 7} \] Next, we swap $ x $ and $ y $: \[ x = \frac{7y - 1}{3y + 7} \] Now, we solve for $ y $: \[ x(3y + 7) = 7y - 1 \] Expanding and rearranging gives: \[ 3xy + 7x = 7y - 1 \implies 3xy - 7y = -7x - 1 \] Factoring out $ y $: \[ y(3x - 7) = -7x - 1 \implies y = \frac{-7x - 1}{3x - 7} \] Thus, the inverse function is: \[ h^{-1}(x) = \frac{-7x - 1}{3x - 7} \]

Step 2: Determine the Domain of $ h^{-1} $

The domain of $ h^{-1}(x) $ is determined by the values of $ x $ that do not make the denominator zero. The denominator is $ 3x - 7 $, so we set: \[ 3x - 7 \neq 0 \implies x \neq \frac{7}{3} \] Thus, the domain of $ h^{-1} $ in interval notation is: \[ (-\infty, \frac{7}{3}) \cup (\frac{7}{3}, \infty) \]

Step 3: Determine the Range of $ h^{-1} $

The range of $ h^{-1}(x) $ corresponds to the domain of the original function $ h(x) $. The denominator of $ h(x) $ is $ 3x + 7 $, which cannot be zero: \[ 3x + 7 \neq 0 \implies x \neq -\frac{7}{3} \] Thus, the range of $ h^{-1} $ in interval notation is: \[ (-\infty, -\frac{7}{3}) \cup (-\frac{7}{3}, \infty) \]

Final Answer

The inverse function is: \[ \boxed{h^{-1}(x) = \frac{-7x - 1}{3x - 7}} \] The domain of $ h^{-1} $ is: \[ \boxed{(-\infty, \frac{7}{3}) \cup (\frac{7}{3}, \infty)} \] The range of $ h^{-1} $ is: \[ \boxed{(-\infty, -\frac{7}{3}) \cup (-\frac{7}{3}, \infty)} \]

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