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Imvt and lmvt .

  1. Mar 25, 2015 #1
    Let f:[0,1]→[0,1] be a continuous function . Show that there exists a point such that f(x) = x
     
  2. jcsd
  3. Mar 25, 2015 #2

    HallsofIvy

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    That's a fairly standard homework exercise. What have you tried?
     
  4. Mar 25, 2015 #3
    Tell me if this arguement is right...... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.
     
  5. Mar 25, 2015 #4
    Apply MVT.
    According to MVT, there exists a point c such that, f'(c)={f(1)-f(0)}/(1-0)=1-0=1
    Now dy/dx=1. So you can find y.
    now substitute a point in the domain to get the value of integration constant.
     
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