showing that a continuous function attains each of its values(or each real number in general) exactly 3 times.

Asked Oct 29 '18 at 05:31

Active Oct 29 '18 at 07:29

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The question and its answer is given below:

But I am wondering how I can prove that the function $f$ attains each real number exactly 3 times, could anyone show me how can I do this?

I also did not understand the proof given in the last paragraph, could anyone explain it for me please?

edited Oct 29 '18 at 07:29

asked Oct 29 '18 at 05:31

Intuition

. See here =https://math.stackexchange.com/q/735842/377379 – RAM_3R Oct 29 '18 at 06:58
@RAM_3R I have looked at it before posting my question ..... it does not contain the answer to my question. – Intuition Oct 29 '18 at 07:06
3

There is a typo: It should be $g_1(x)=2-x$ for $1\leq x\leq 2$. – Christian Blatter Oct 29 '18 at 07:26
@ChristianBlatter why it is a typo? – Intuition Oct 29 '18 at 07:49
See related https://math.stackexchange.com/q/677085/72031 – Paramanand Singh Oct 29 '18 at 08:14
@ParamanandSingh is the given example above contains a typo? – Intuition Oct 29 '18 at 08:17
Yes there is a typo as the given $g_1$ is discontinuous at $x=1$ – Paramanand Singh Oct 29 '18 at 08:30
Is there any other Typos? – Intuition Oct 29 '18 at 08:36
1

Draw the graph of this function to see why it takes each value exactly thrice. – DanielWainfleet Oct 29 '18 at 17:26

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