Questions: Given the function p(x)=12/(4x+11), determine the inverse function p^(-1)(x) in simplified form: p^(-1)(x)= Determine the domain and range for both p(x) and p^(-1)(x) using interval notation: Domain of p(x) : Domain of p^(-1)(x) : Range of p(x) : Range of p^(-1)(x) :

Given the function p(x)=12/(4x+11), determine the inverse function p^(-1)(x) in simplified form:
p^(-1)(x)=

Determine the domain and range for both p(x) and p^(-1)(x) using interval notation:
Domain of p(x) :

Domain of p^(-1)(x) :

Range of p(x) :

Range of p^(-1)(x) :

Transcript text: 12. Given the function $p(x)=\frac{12}{4 x+11}$, determine the inverse function $p^{-1}(x)$ in simplified form: \[ p^{-1}(x)= \] Determine the domain and range for both $p(x)$ and $p^{-1}(x)$ using interval notation: Domain of $p(x)$ : Domain of $p^{-1}(x)$ : Range of $p(x)$ : Range of $p^{-1}(x)$ :

Solution

Solution Steps

To find the inverse function $ p^{-1}(x) $ of $ p(x) = \frac{12}{4x + 11} $, we need to solve for $ x $ in terms of $ y $ where $ y = p(x) $. Then, we will express $ x $ as a function of $ y $. After finding the inverse function, we will determine the domain and range of both $ p(x) $ and $ p^{-1}(x) $.

Solution Approach

Start with $ y = \frac{12}{4x + 11} $.
Solve for $ x $ in terms of $ y $.
Express $ x $ as a function of $ y $ to get $ p^{-1}(x) $.
Determine the domain and range of $ p(x) $ and $ p^{-1}(x) $.

Step 1: Finding the Inverse Function

To find the inverse function $ p^{-1}(x) $ of $ p(x) = \frac{12}{4x + 11} $, we start with the equation: \[ y = \frac{12}{4x + 11} \] Rearranging this equation to solve for $ x $ gives: \[ x = -\frac{11}{4} + \frac{3}{y} \] Thus, the inverse function is: \[ p^{-1}(x) = -\frac{11}{4} + \frac{3}{x} \]

Step 2: Domain of $ p(x) $

The function $ p(x) $ is undefined when the denominator equals zero: \[ 4x + 11 = 0 \implies x = -\frac{11}{4} \approx -2.75 \] Therefore, the domain of $ p(x) $ is: \[ \text{Domain of } p(x): \quad x \neq -\frac{11}{4} \]

Step 3: Range of $ p(x) $

The range of $ p(x) $ excludes the value where the function approaches zero: \[ \text{Range of } p(x): \quad y \neq 0 \]

Step 4: Domain of $ p^{-1}(x) $

The domain of the inverse function $ p^{-1}(x) $ corresponds to the range of $ p(x) $: \[ \text{Domain of } p^{-1}(x): \quad x \neq 0 \]

Step 5: Range of $ p^{-1}(x) $

The range of the inverse function $ p^{-1}(x) $ corresponds to the domain of $ p(x) $: \[ \text{Range of } p^{-1}(x): \quad y \neq -\frac{11}{4} \]

Final Answer

\[ p^{-1}(x) = -\frac{11}{4} + \frac{3}{x} \] \[ \text{Domain of } p(x): \quad x \neq -\frac{11}{4} \] \[ \text{Range of } p(x): \quad y \neq 0 \] \[ \text{Domain of } p^{-1}(x): \quad x \neq 0 \] \[ \text{Range of } p^{-1}(x): \quad y \neq -\frac{11}{4} \]

Thus, the final boxed answers are: \[ \boxed{p^{-1}(x) = -\frac{11}{4} + \frac{3}{x}} \] \[ \boxed{\text{Domain of } p(x): x \neq -\frac{11}{4}} \] \[ \boxed{\text{Range of } p(x): y \neq 0} \] \[ \boxed{\text{Domain of } p^{-1}(x): x \neq 0} \] \[ \boxed{\text{Range of } p^{-1}(x): y \neq -\frac{11}{4}} \]

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