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Prove that ∃ a∈]0,2[ such that f(a)=a^2

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Let f:[0,2] --> ]0,4[ be continuous function. Prove that ∃ a∈]0,2[ such that f(a)=a2

    2. Relevant equations


    3. The attempt at a solution
    Sadly I have no idea whatsoever.
    if e.g. a=1 and f(x)=x, then f(1)=12
     
  2. jcsd
  3. Oct 4, 2015 #2

    HallsofIvy

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    Look at the function g(x)= f(x)- x^2. g(0)= f(0) and g(2)= f(2)- 4. Since the largest possible value for f is 4, what can you say about g(2)?
     
  4. Oct 4, 2015 #3

    Krylov

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    Apply the intermediate value theorem to a suitably chosen function ##g: [0,2] \to [0,4]##.
     
  5. Oct 6, 2015 #4
    I still don't get it :(
     
  6. Oct 6, 2015 #5

    Krylov

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    Is there something specific that you don't get? Did you make a start?

    Could you perhaps give a precise formulate of the intermediate value theorem, as you understand it? (I suppose this question comes from a calculus course, where this theorem should appear.)
     
  7. Oct 6, 2015 #6

    HallsofIvy

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    You were told to look at the function [itex]g(x)= f(x)- x^2[/itex]. Now you have been told to use the "intermediate value theorem". Do you know what that is?
     
  8. Oct 6, 2015 #7
    I don't know how to get started and how to apply the intermediate value theorem.

    Intermediate value theorem: If function f is continuous in [a,b] and f(a)<u<f(b), then there exists c ∈]a,b[ such that f(c)=u.

    f(0)< a2 <f(2) --> a∈]0,2[ such that f(a)=a2
     
    Last edited: Oct 6, 2015
  9. Oct 6, 2015 #8

    HallsofIvy

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    And what have I already told you, last Sunday, about g(0) and g(4)?
     
  10. Oct 6, 2015 #9

    Krylov

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    Careful, the function ##f## in your statement of the IVT is not the same as the function ##f## from your exercise. Replace ##f## in the IVT by ##g## as proposed earlier. Also, the IVT is true as well when we replace "##f(a) < u < f(b)##" by "##f(a) > u > f(b)##".
     
  11. Oct 6, 2015 #10
    ....
     
    Last edited: Oct 6, 2015
  12. Oct 6, 2015 #11
    You want to find a zero of the function ##g(x) = f(x) - x^2 ##.
    If you prove there is ##x_0## and ##x_1## in ##[0,2]## such that ##g(x_0) < 0 ## and ##g(x_1) > 0 ##, then the intermediate value theorem guarantees the existence of a zero for ##g##.

    You have ## 0 < f < 4 ##. Then what about ##g(0)## and ##g(2)## ?
     
    Last edited: Oct 6, 2015
  13. Oct 6, 2015 #12
    g(0)>0 and g(2)<0
    g(0)>0>g(2) so IVT: ∃x0∈]0,2[ such that g(x0)=0 ⇔ f(x0)=x02
    Now I got it, I appreciate your help! thanks everyone!
     
  14. Oct 6, 2015 #13
     
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