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Mean Value Theorem - Find the values of c

  1. Nov 9, 2009 #1
    1. Verify that the function satisfies the mean value theorem, then find all numbers c that satisfy the the conclusion of the mvt

    f(x)=e^(-2x) on the interval [0,3]

    2. f'(c)=[f(b)-f(a)]/[b-a]



    3.
    1. f(x) is a composition of continuous functions, so f(x) is continuous on[0,3].
    2. f(x) is a composition of differentiable functions, so it is differentiable on (0,3).

    so, f(b)-f(a)=f'(c)(b-a)
    e^-6-e^0=(-2*e^-2c)(3)

    Then take the ln of both sides...
    -6-0=ln(-6*e^-2c)
    -6=ln(-6)+2c

    Solve for c..?
    c=[-6-ln(-6)]/[-2]

    Can't have ln of a negative..(well without imaginary numbers). What's gone wrong? I'm stumped.
     
  2. jcsd
  3. Nov 9, 2009 #2

    LCKurtz

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    You have
    e-6 - e0=(-2*e-2c)(3) = -6e-2c

    At that point both sides of the equation are negative. So multiply both sides by -1 before taking logarithms.
     
  4. Nov 10, 2009 #3

    HallsofIvy

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    NO. ln(e^{-6}- e^0) is NOT -6-0. In general, ln(a+b) is NOT ln(a)+ ln(b).

     
  5. Nov 10, 2009 #4
    Hehehe... I had my value for b-a incorrect, which threw off my whole solution.. :D Got the right answer now, thanks for the feedback ;)
     
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