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Mean Value Theorem answer help

  1. Apr 7, 2009 #1
    1. If c is the value defined by the mean value theorem, then for f(x) = e^x - x^2 on [0,1], c=

    I found the two end points as [0,1] and [1,e-1], so the average slope is .71828...

    is that the answer then?
     
  2. jcsd
  3. Apr 7, 2009 #2
    nope, whats the formula for the MVT?
     
  4. Apr 7, 2009 #3
    f(b)-f(a) / b-a

    so (e-1) - (1-1) / (1 - 0)

    Which gives 1.718...
     
  5. Apr 7, 2009 #4
    How does your book state the mean value theorem? This is important. My book states that if f is continuous on [a,b] and f is differentiable on (a,b), then there exists a number c in (a,b) such that
    [tex] f'(c) = \frac{f(b) - f(a)}{b-a} [/itex]

    You found the quotient
    [tex] \frac{f(1) - f(0)}{1-0} = e-2[/tex]
    (Don't put it in decimal form.)

    Now if your book states the mean value theorem this way (which it most likely does), then you need to find the c in (a,b) such that f'(c)=e-2, which means you have an incorrect answer.
     
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