Question 1162522
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(f\,\circ\,g\)(x)\ =\ f\(g(x)\)\ =\ \log\(g(x)\)\ =\ \log\(\frac{1}{x\,+\,3}\)]


Since the domain of the log function is *[tex \Large \mathbb{R}^+], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x\,+\,3}\ >\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 3\ >\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ -3]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{Dom }\(f\,\circ\,g\)\ =\ \{x\,\in\,\mathbb{R}\,|\,x\,>\,-3\}]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(g\,\circ\,f\)(x)\ =\ g\(f(x)\)\ =\ \frac{1}{f(x)\,+\,3}\ =\ \frac{1}{\log(x)\,+\,3}]


Since rational functions are not defined for any value of the independent variable that causes the denominator of the function to equal zero,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x)\,+\,3\ \neq\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(x)\ \neq\ -3]


So, depending on whether you mean *[tex \Large \log_{10}] or *[tex \Large \ln] when you write *[tex \Large \log] without specifying the base,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{Dom }\(g\,\circ\,f\)\ =\ \{x\,\in\,\mathbb{R}\,|\,x\,\neq\,10^{-3}\}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{Dom }\(g\,\circ\,f\)\ =\ \{x\,\in\,\mathbb{R}\,|\,x\,\neq\,e^{-3}\}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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