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Given the function: $F(x)=x \sin(\frac{1}{x})$ when $x \not= 0$, $0$ when $x=0$ We proved that this function continuous. The question that raises: given a certain interval $[a,b]$ when a,b are real numbers, is there such a point $c$ where in the interval $[a,c]$ the function $F$ is monotonic?

I think that the answer is yes, my method to show it is: given a point d in $[a,b]$. we look at the interval $[a,d]$, the interval is continuous. in the interval $[a,d]$ we at all the values of the function, is there one that is repeated twice? (injective) if so, we look at $[a,\frac{d}{2}]$. we ask the same question- if the answer is yet no then we repeat. I think that we will get to an answer because otherwise we will contradict the fact that $[F(a),F(b)]$ is continuous.

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    What does it mean for an interval to be continuous? – 2017-01-18
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    https://en.wikipedia.org/wiki/Interval_(mathematics) interval i mean the group of values [a,b], continuos interval is an interval that includes all the values in the range. when a value isn't in the interval for example we might have https://en.wikipedia.org/wiki/Classification_of_discontinuities – 2017-01-18
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    "interval is continuous" is not defined there. – 2017-01-18
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    Just say "interval". That has all the information you need. – 2017-01-18

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Hint: In any interval $[0,c],$ the derivative of this function will take on positive and negative values.