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Maximum, minimum, and continuity

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap2.jpg
    Theorem 3 that i will give in the attempt at a solution talks about a closed interval, here it's open
    2. Relevant equations
    Continuity:
    $$\vert x-c \vert < \delta~\Rightarrow~\vert f(x)-f(c) \vert < \epsilon$$
    $$\delta=\delta(c,\epsilon)$$

    3. The attempt at a solution
    Snap3.jpg
     
  2. jcsd
  3. Jun 2, 2017 #2

    Mark44

    Staff: Mentor

    Theorem 3 is no help here, since the interval for this problem is 0 < x < 1. The problem is fairly simple -- you shouldn't need to invoke a theorem to answer it.
     
  4. Jun 2, 2017 #3
    The derivative 2x is horizontal at 0, so there is a minimum or maximum. the fact that there isn't any other zero derivative proves there isn't another bending and it's rising, so the maximum is at 1
     
  5. Jun 2, 2017 #4
    Remember ##0 < x < 1##. Also you don't need to think this in terms of derivatives.
     
  6. Jun 2, 2017 #5

    LCKurtz

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    ##x=0## and ##x=1## are not in the domain of your function.
     
  7. Jun 2, 2017 #6

    Ray Vickson

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    The points x=0 and x=1 are not in the domain of the given function!
     
  8. Jun 2, 2017 #7

    pasmith

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    Hint: What is the distinction between a maximum and a supremum?
     
  9. Jun 2, 2017 #8
    1 is the supremum, x2 has no maximum.
    0 is the infinum, x2 has no minimum
    Is it the answer? it isn't part of the chapter, i learned about supremum here
     
  10. Jun 2, 2017 #9

    Mark44

    Staff: Mentor

    Colored text added.
     
  11. Jun 2, 2017 #10
    So is it the answer?
     
  12. Jun 2, 2017 #11

    Ray Vickson

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    Already answered!
     
  13. Jun 2, 2017 #12
    Thank you very much Ray, Mark, pasmith, LCKurtz and dgambh
     
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