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Find h(x, y) = g(f(x, y)). h(x, y) = g(t) = t + In(t), f(x, y) = 4-xy 5+x²y² +In 4 - xy 5 + x²y² X 4-xy 5+x²y² Find the set of points at which h is continuous. OD = {(x, y) | xy <4} OD = {(x, y) | xy ≤ 4} Oh is continuous on R2 OD = {(x, y) |xy > 4} OD = {(x, y) | xy ≥ 4}

Find h(x, y) = g(f(x, y)). h(x, y) = g(t) = t + In(t), f(x, y) = 4-xy 5+x²y² +In 4 - xy 5 + x²y² X 4-xy 5+x²y² Find the set of points at which h is continuous. OD = {(x, y) | xy <4} OD = {(x, y) | xy ≤ 4} Oh is continuous on R2 OD = {(x, y) |xy > 4} OD = {(x, y) | xy ≥ 4}

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Explanation on Finding Continuity for Composite Functions In this example, we are given the task of finding \( h(x,y) = g(f(x,y)) \). - **Function Definitions:** - \( g(t) = t + \ln(t) \) - \( f(x,y) = \frac{4 - xy}{5 + x^2 y^2} \) - **Composite Function:** - \( h(x,y) = \frac{4 - xy}{5 + x^2 y^2} + \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) ### Determining Continuity Points To determine the set of points at which \( h \) is continuous, we need to evaluate the points where the composite function \( h(x,y) \) avoids undefined values and discontinuities. ### Evaluating Discontinuities: The function \( h(x,y) \) can be divided into two parts: 1. \( \frac{4 - xy}{5 + x^2 y^2} \) 2. \( \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) For the logarithm \( \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) to be defined, the argument inside the logarithm must be positive: \[ \frac{4 - xy}{5 + x^2 y^2} > 0 \] This requires \( 4 - xy > 0 \) because the denominator \( 5 + x^2 y^2 > 0 \) is always positive for all \( x \) and \( y \) in \( \mathbb{R} \). Therefore: \[ |xy| < 4 \] ### Conclusion: The correct set of points at which the function \( h \) is continuous is: \[ D = \{(x, y) \mid |xy| \leq 4\} \] This is marked with a radio button selection in the provided image: - Option: \( D = \{(x, y) \mid |xy| \leq 4\} \) has been selected as the correct answer. This selection is indicated by a blue circle around the option and a green checkmark next to
Transcribed Image Text:### Explanation on Finding Continuity for Composite Functions In this example, we are given the task of finding \( h(x,y) = g(f(x,y)) \). - **Function Definitions:** - \( g(t) = t + \ln(t) \) - \( f(x,y) = \frac{4 - xy}{5 + x^2 y^2} \) - **Composite Function:** - \( h(x,y) = \frac{4 - xy}{5 + x^2 y^2} + \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) ### Determining Continuity Points To determine the set of points at which \( h \) is continuous, we need to evaluate the points where the composite function \( h(x,y) \) avoids undefined values and discontinuities. ### Evaluating Discontinuities: The function \( h(x,y) \) can be divided into two parts: 1. \( \frac{4 - xy}{5 + x^2 y^2} \) 2. \( \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) For the logarithm \( \ln \left( \frac{4 - xy}{5 + x^2 y^2} \right) \) to be defined, the argument inside the logarithm must be positive: \[ \frac{4 - xy}{5 + x^2 y^2} > 0 \] This requires \( 4 - xy > 0 \) because the denominator \( 5 + x^2 y^2 > 0 \) is always positive for all \( x \) and \( y \) in \( \mathbb{R} \). Therefore: \[ |xy| < 4 \] ### Conclusion: The correct set of points at which the function \( h \) is continuous is: \[ D = \{(x, y) \mid |xy| \leq 4\} \] This is marked with a radio button selection in the provided image: - Option: \( D = \{(x, y) \mid |xy| \leq 4\} \) has been selected as the correct answer. This selection is indicated by a blue circle around the option and a green checkmark next to
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