Recently this problem was given to me by a friend. I am unable to understand how to approach this. \(\textbf{NOTE}\): I do not know about the validity of the problem.

If \(f(x)=sinx\) has exactly \(2\) solutions, how many solutions can \(f(\frac{1}{x})=sinx\) have? \([f(x):\mathbb{R}\to\mathbb{R}]\)

The answer given is \(\boxed{3}\)

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## Comments

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TopNewestGiven only the information provided, my answer would be "arbitrarily many". Are you missing out some conditions?

E.g. Is the function defined from the reals to the reals? Must the function be continuous? differentiable? strictly increasing? etc.

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yes, i am sorry, the function is \(\text{ is continuous and differentiable}\) and is defined from \(:\mathbb{R}\to\mathbb{R}\) again, i am extremely sorry for any confusion over this subject

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Even under those assumptions, my answer is still "arbitrarily many".

Think about what happens if \( f(x) = 0 \) in a neighborhood about \(0\).

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