No. of solutions

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


If f(x)=sinxf(x)=sinx has exactly 22 solutions, how many solutions can f(1x)=sinxf(\frac{1}{x})=sinx have? [f(x):RR][f(x):\mathbb{R}\to\mathbb{R}]

The answer given is 3\boxed{3}

#Calculus #Functions #Solutions

Note by Aritra Jana
6 years ago

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1 vote

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Comments

Given 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.

Calvin Lin Staff - 6 years ago

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

Aritra Jana - 6 years ago

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

Think about what happens if f(x)=0 f(x) = 0 in a neighborhood about 00.

Calvin Lin Staff - 6 years ago
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