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I was going through the topic of $Function$, its boundedness, continuity etc. I got a problem.

Does there exist a function defined on the closed interval $[a,b]$ which is....

1. bounded;

2. takes its maximum and minimum values;

3. takes all its values between the maximum and minimum values;

Then can we conclude that then this function is continuous at some points or subintervals on $[a,b]$.

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Function below satisfies all three conditions above but it is discontinuous at every point on $[-1,1]$

$f(x)=\begin{cases} 1,&\text{if $x = 0$}\\ x,&\text{if x is rational, $x \neq 0$ , $x\neq 1$}\\ -x,&\text{if x is rational, $x \neq 0$ , $x\neq 1$, $x\neq -1$} \\ 0,&\text{if x = 1} \end{cases}$

It is impossible to draw the graph of the function $y = f(x)$ but the sketch below gives an idea of its behavior.

enter image description here

Hence answer is no.

You can find this example in counter examples in calculus.

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No. Take $f:[0,1]\to[0,1]$ such that $f(x) = x$ if $x$ is rational and $f(x)=1-x$ if $x$ is irrational. Well, okay, that's continuous at $x=\frac{1}{2}$, so just set $f(0) = \frac{1}{2}$ and $f(1/2) = 0$ and use the rule I gave above for all other values of $x$. :)

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