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For questions 5 through 10 , consider \( m(x)=\frac{2}{x}, t(x)=2 x^{2}+3 x, w(x)=2 x+5 \). 5. Determine \( (m \circ t)(2) \). 6. Determine \( (t \circ w)(x) \). 7. Determine the domain of \( (t \circ w)(x) \). Determine \( (t \circ m)(x) \). Determine the domain of \( (t \circ m)(x) \).

Ask by Barker Mcfarlane. in the United States
Jan 21,2025

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Answer

\[ (m \circ t)(2) = \frac{1}{7} \] \[ (t \circ w)(x) = 8x^2 + 46x + 65 \] \[ \text{Domain of } (t \circ w)(x) = \mathbb{R} \] \[ (t \circ m)(x) = \frac{8}{x^2} + \frac{6}{x} \] \[ \text{Domain of } (t \circ m)(x) = \mathbb{R} \setminus \{0\} \]

Solution

Let's address each of the questions step by step using the given functions: - \( m(x) = \frac{2}{x} \) - \( t(x) = 2x^2 + 3x \) - \( w(x) = 2x + 5 \) --- ### **5. Determine \( (m \circ t)(2) \)** **Definition:** \[ (m \circ t)(x) = m(t(x)) \] **Calculation:** 1. **Find \( t(2) \):** \[ t(2) = 2(2)^2 + 3(2) = 2(4) + 6 = 8 + 6 = 14 \] 2. **Find \( m(t(2)) = m(14) \):** \[ m(14) = \frac{2}{14} = \frac{1}{7} \] **Answer:** \[ (m \circ t)(2) = \frac{1}{7} \] --- ### **6. Determine \( (t \circ w)(x) \)** **Definition:** \[ (t \circ w)(x) = t(w(x)) \] **Calculation:** 1. **Express \( t(w(x)) \):** \[ t(w(x)) = 2(w(x))^2 + 3w(x) = 2(2x + 5)^2 + 3(2x + 5) \] 2. **Expand and simplify:** \[ (2x + 5)^2 = 4x^2 + 20x + 25 \] \[ t(w(x)) = 2(4x^2 + 20x + 25) + 3(2x + 5) = 8x^2 + 40x + 50 + 6x + 15 = 8x^2 + 46x + 65 \] **Answer:** \[ (t \circ w)(x) = 8x^2 + 46x + 65 \] --- ### **7. Determine the Domain of \( (t \circ w)(x) \)** **Analysis:** - **Function \( w(x) = 2x + 5 \):** Defined for all real numbers. - **Function \( t(x) = 2x^2 + 3x \):** Also defined for all real numbers since it's a polynomial. **Conclusion:** Since both \( w(x) \) and \( t(x) \) are defined for all real numbers, their composition \( t(w(x)) \) is also defined for all real numbers. **Answer:** \[ \text{Domain of } (t \circ w)(x) = \mathbb{R} \quad \left(-\infty,\, \infty\right) \] --- ### **8. Determine \( (t \circ m)(x) \)** **Definition:** \[ (t \circ m)(x) = t(m(x)) \] **Calculation:** 1. **Express \( m(x) \):** \[ m(x) = \frac{2}{x} \] 2. **Substitute into \( t(x) \):** \[ t\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 + 3\left(\frac{2}{x}\right) = 2\left(\frac{4}{x^2}\right) + \frac{6}{x} = \frac{8}{x^2} + \frac{6}{x} \] **Simplified Form:** \[ (t \circ m)(x) = \frac{8}{x^2} + \frac{6}{x} \quad \text{or} \quad \frac{8 + 6x}{x^2} \] **Answer:** \[ (t \circ m)(x) = \frac{8}{x^2} + \frac{6}{x} \] --- ### **9. Determine the Domain of \( (t \circ m)(x) \)** **Analysis:** - **Function \( m(x) = \frac{2}{x} \):** Defined for all \( x \) except \( x = 0 \). - **Function \( t(x) = 2x^2 + 3x \):** Defined for all real numbers. **Composition \( t(m(x)) \):** - The inner function \( m(x) \) must be defined, which requires \( x \neq 0 \). - Since \( t(x) \) is defined for all real numbers, no additional restrictions are imposed. **Conclusion:** The domain of \( (t \circ m)(x) \) consists of all real numbers except \( x = 0 \). **Answer:** \[ \text{Domain of } (t \circ m)(x) = \mathbb{R} \setminus \{0\} \quad \left(-\infty,\, 0\right) \cup \left(0,\, \infty\right) \] --- Feel free to ask if you need further clarifications or additional assistance!

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Let’s tackle each of the questions step by step. ### 5. Determine \( (m \circ t)(2) \). To find \( (m \circ t)(2) \), we first need to evaluate \( t(2) \): \[ t(2) = 2(2^2) + 3(2) = 2(4) + 6 = 8 + 6 = 14. \] Now we apply \( m \) to this result: \[ m(t(2)) = m(14) = \frac{2}{14} = \frac{1}{7}. \] Thus, \( (m \circ t)(2) = \frac{1}{7} \). ### 6. Determine \( (t \circ w)(x) \). To find \( (t \circ w)(x) \), we substitute \( w(x) \) into \( t(x) \): \[ w(x) = 2x + 5, \] Now substitute \( w(x) \) into \( t(x) \): \[ t(w(x)) = t(2x + 5) = 2(2x + 5)^2 + 3(2x + 5). \] Expanding \( (2x + 5)^2 \): \[ (2x + 5)^2 = 4x^2 + 20x + 25, \] So: \[ t(w(x)) = 2(4x^2 + 20x + 25) + 3(2x + 5) = 8x^2 + 40x + 50 + 6x + 15 = 8x^2 + 46x + 65. \] Thus, \( (t \circ w)(x) = 8x^2 + 46x + 65 \). ### 7. Determine the domain of \( (t \circ w)(x) \). The function \( w(x) = 2x + 5 \) is defined for all real numbers. Since \( t(x) = 2x^2 + 3x \) is also a polynomial, it is defined for all real numbers as well. Therefore, the domain of \( (t \circ w)(x) \) is all real numbers: \[ \text{Domain of } (t \circ w)(x) = (-\infty, \infty). \] ### 8. Determine \( (t \circ m)(x) \). To find \( (t \circ m)(x) \), we substitute \( m(x) \) into \( t(x) \): \[ m(x) = \frac{2}{x}, \] Now, substitute \( m(x) \) into \( t(x) \): \[ t(m(x)) = t\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 + 3\left(\frac{2}{x}\right). \] This expands to: \[ t\left(\frac{2}{x}\right) = 2 \cdot \frac{4}{x^2} + \frac{6}{x} = \frac{8}{x^2} + \frac{6}{x}. \] Thus, \( (t \circ m)(x) = \frac{8}{x^2} + \frac{6}{x} \). ### 9. Determine the domain of \( (t \circ m)(x) \). The function \( m(x) = \frac{2}{x} \) is defined for all real numbers except \( x = 0 \). Hence, the output of \( m(x) \) will also exclude 0. Therefore, since \( t(x) \) takes all values except for 0 should be considered: \[ \text{Domain of } (t \circ m)(x) = x \in \mathbb{R}, x \neq 0 \to (-\infty, 0) \cup (0, \infty). \]

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