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Math Help - Max -Min theorm ( please help me to solve this problem )

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    Max -Min theorm ( please help me to solve this problem )

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    Senior Member Sambit's Avatar
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    For the first problem, the function will be continuous if f(2+)=f(2-)=f(2) where 2+ means some value of x which is just larger than 2 and 2- means some value of x which is just smaller than 2. So obtain the value of f(2+)=f(2-) from f(x) and substitute K by that value. Understood?
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    if f(2+)= 20
    f(2-) = 20
    f(2)= k = f(2+) = f(2-)= 20 Is this right ?????

    but how can i decide if the function f(x) given in a) have a maxima or a
    minima on the intervalls mentioned in the question ?????????????????
    Max-Min theorem says :
    if f(x) is cont. on the closed interval [a,b], then there exist numbers p,q in the interval [a,b] such that for all x in [a,b]
    f(p) less than or equal to f(x) less that or equal to f(q)
    Thus f has the absolute min. value m=f(p) taken on at the point p , and the absolute max. value M =f(q)
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    Senior Member Sambit's Avatar
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    Yes you are right for the first part.
    For the second part, I am not very much aware of that theorem,though I think f(x)=5x^2 in case x is not equal to 2, so max is obtained at x=4 and minimum at x=0.
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    Thanks alot for replying
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Sambit View Post
    For the second part, I am not very much aware of that theorem,though I think f(x)=5x^2 in case x is not equal to 2, so max is obtained at x=4 and minimum at x=0.

    The function f(x)=5x^2 is strictly increasing for x\geq 0 so we have:

    (i) On [0,4] absolute minimum at x=0 and absolute maximum at x=4 .

    (ii) On (0,4) no absolute minimum and no absolute maximum .

    (iii) On (0,4] no absolute minimum and absolute maximum at x=4 .


    Fernando Revilla
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    Note, however, that the problem does NOT ask you to determine where the max and min occur, only if they exist.
    For (i), at least, it is sufficient to say that the function is continuous on that closed and bounded interval and so both max and min exist.

    (ii) and (iii) are a little harder since the max-min theorem does not say what happens if the interval is not closed.
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    Quote Originally Posted by FernandoRevilla View Post
    The function f(x)=5x^2 is strictly increasing for x\geq 0 so we have:

    (i) On [0,4] absolute minimum at x=0 and absolute maximum at x=4 .

    (ii) On (0,4) no absolute minimum and no absolute maximum .

    (iii) On (0,4] no absolute minimum and absolute maximum at x=4 .


    Fernando Revilla


    Thank you very much
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mariama View Post
    Thanks

    Don't waste a post. Better press the Thanks button.


    Fernando Revilla
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