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Math Help - Help with proving the example

  1. #1
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    Help with proving the example

    Provide an example of a continuous function f:R-->R which satisfies all of the following properties

    (a) non-decreasing
    (b) not constant
    (c) not increasing

    Be sure to justify your example

    I have an example as f( x ) = { 1, x<0 or x, x >= 0 with the graph that the function is constant on the left of the x axis, and increasing on the right of the x axis, but I dont know how to prove this
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  2. #2
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    Quote Originally Posted by 450081592 View Post
    Provide an example of a continuous function f:R-->R which satisfies all of the following properties

    (a) non-decreasing
    (b) not constant
    (c) not increasing

    Be sure to justify your example

    I have an example as f( x ) = { 1, x<0 or x, x >= 0 with the graph that the function is constant on the left of the x axis, and increasing on the right of the x axis, but I dont know how to prove this
    \sin x\,,\,\,\cos x\,,\,\,x^2\,,\,\,etc..

    Your example is also called weakly increasing since x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2) , so it may not qualify as a right example.

    Tonio
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    Quote Originally Posted by tonio View Post
    \sin x\,,\,\,\cos x\,,\,\,x^2\,,\,\,etc..

    Your example is also called weakly increasing since x_1<x_2\,\Longrightarrow\,f(x_1)\leq f(x_2) , so it may not qualify as a right example.

    Tonio
    sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions
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    Quote Originally Posted by 450081592 View Post
    sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions
    No, a function is non-decreasing if for ALL x_1 < x_2 \in \mathbb{R} , \ f(x_1) \leq f(x_2) -- not for SOME x_1,x_2.. tonio's answers are perfectly fine.
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    Quote Originally Posted by 450081592 View Post
    sinx con x and x^2 won't work cause they decrease somewhere, the function should never decrease in order to sasitify the conditions

    There's no such a thing! If a function is not constant then there are points  x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y) !
    ANY non-constant function will increase/decrease "somewhere"...

    Tonio
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    Quote Originally Posted by tonio View Post
    There's no such a thing! If a function is not constant then there are points  x < y\,\, s.t.\,\, f(x) <f(y)\,\,or\,\,f(x)>f(y) !
    ANY non-constant function will increase/decrease "somewhere"...

    Tonio
    not decrease is not the same as non-decrease, non-decrease means the function never decreases, not decrease means the function not always decrease, that what my prof said
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    Quote Originally Posted by 450081592 View Post
    not decrease is not the same as non-decrease, non-decrease means the function never decreases, not decrease means the function not always decrease, that what my prof said

    Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

    Tonio
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    Quote Originally Posted by tonio View Post
    Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

    Tonio
    This is not obvius.
    Consider the function:

    f(x)=1 if x\in\mathbb{Q}, and f(x)=0 if x\in\mathbb{R}\setminus\mathbb{Q}

    Now there exist points x_1<x_2 such that f(x_1)<f(x_2),
    but the function is still not increasing in [x_1,x_2] no matter what points you pick.

    Weierstrass function - Wikipedia, the free encyclopedia is a continuous function with the same attribute. There is no interval where the function is increasing or decreasing.
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  9. #9
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    Quote Originally Posted by tonio View Post
    Whatever: there is no such a thing like a non-constant function that never decreases and never increases: it just is an oxymoron, and the reason was given in my last post.

    Tonio
    nope, what the 3 conditions mean
    (a) non-decreasing means never decrease, it can never decrease
    (b) not constant means not always constant, so it can be contant in a subinterval
    (c) not increasing means not always increase, so it can increase in a subinterval
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    Quote Originally Posted by hjortur View Post
    This is not obvius.
    Consider the function:

    f(x)=1 if x\in\mathbb{Q}, and f(x)=0 if x\in\mathbb{R}\setminus\mathbb{Q}

    Now there exist points x_1<x_2 such that f(x_1)<f(x_2),
    but the function is still not increasing in [x_1,x_2] no matter what points you pick.

    Weierstrass function - Wikipedia, the free encyclopedia is a continuous function with the same attribute. There is no interval where the function is increasing or decreasing.
    The only one that mentioned "interval" that I'm aware of was Defunkt trying to explain my answer to the OP according to what he wrote. And in your example 0<\sqrt{2}\,\,and\,\,f(0)>f(\sqrt{2}).
    Perhaps the OP meant that there is no INTERVAL where the function increases or decreases and then yours or Wierstrass' function work fine.
    It never minds though, since the OP knows better because "his professor told him".

    Tonio
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