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Q24E

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Essential Calculus: Early Transcendentals
Found in: Page 632
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Determine the set of points at which the function is continuous.

\(H\left( {x,y} \right) = \frac{{{e^x} + {e^y}}}{{{e^{xy}}{\rm{ - }}1}}\)

The set of points for which the function is continuous is

\(\left\{ {\left( {x,y} \right)\left| {{e^{xy}}{\rm{ - }}1 \ne 0} \right|} \right\}\)

See the step by step solution

Step by Step Solution

Step-1: Recognising The Form Of Function

The function is \(H\left( {x,y} \right) = \frac{{{e^x} + {e^y}}}{{{e^{xy}}{\rm{ - }}1}}\)

The function H is a composition of elementary functions. A composition of elementary functions is always continuous on its domain.

The exponential function \({e^x}\)has no restrictions on t.

Therefore, the only restriction on \(H\left( {x,y} \right) = \frac{{{e^x} + {e^y}}}{{{e^{xy}}{\rm{ - }}1}}\)is \({e^{xy}}{\rm{ - }}1 \ne 0\),

(The denominator cannot be 0)

Therefore, the set of points for which the function is continuous is \(\left\{ {\left( {x,y} \right)\left| {{e^{xy}}{\rm{ - }}1 \ne 0} \right|} \right\}\)

This can be simplified further as follows:

\(\begin{aligned}{l}{e^{xy}}{\rm{ - }}1 \ne 0\\ \Rightarrow {e^{xy}} = 1\\ \Rightarrow xy \ne 0\end{aligned}\)

Therefore, the set can be rewritten as

\(\left\{ {\left( {x,y} \right)\left| {xy \ne 0} \right|} \right\}\)

\(\)

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