in the function f(x) = 1 + 1/x, when x= 0 does f(x) remain real? apparently not (seeing the graph), but why? shouldnt f(x) = 1? is my graph wrong?
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Originally Posted by the kopite in the function f(x) = 1 + 1/x, when x= 0 does f(x) remain real? apparently not (seeing the graph), but why? shouldnt f(x) = 1? is my graph wrong? It makes no sense to talk about the value of f(x) when x = 0. Assuming you mean : dom f = R\{0}. x = 0 is a vertical asymptote. ran f = R\{1}. y = 1 is a horizontal asymptote.
which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question.
Originally Posted by the kopite which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question. I told you that it makes no sense to consider the value when x = 0. It is neither real nor not real. It is not defined.
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Suppose This gives Hence there does not exist an such that . EDIT: this is quite apparent since for which cannot happen (since it leads to the same contradiction as before).
Basically, you can never divide by zero. is undefined. It simply does not exist, and there is nothing wrong with your graph.
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