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g(x) = 1/ (1-e^(x-1)) What are the natural domain and range of g? anyone knows how to solve this? thank you
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\[g(x) = \frac{1}{(1-e^{x-1})}\] So, the problem is to determine the domain. Which is another way of saying, what are the values of x which provide me with a real answer. In this case, the denominator can never equal zero. So, solve: \[1-e^{x-1} = 0\] to determine where your function is undefined. \[e^{x-1} = 1\] Which simplifies (by taking ln of each side) \[\ln(e^{x-1}) = \ln(1)\] \[x-1 = \ln(1)\] \[x = \ln(1)+1\] So, when x is equal to ln(1) + 1, my function is undefined. For other values of x, I'm good to go. \[Domain = (-\infty, \ln(1)+1) \cup (\ln()+1),\infty)\]
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