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destinc

  • 3 years ago

Find all numbers c that satisfy the mean value theorem, f(x)=e^2x, [0,3]

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  1. AntiMatter
    • 3 years ago
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    mean value theorem states, that if a function is continuous and differentiable on a given interval [w,z], then there exists a value v, where the derivative of the value of the derivative of the function at v is equal to \[f'(v) = (f(z) - f(w))/(z-w)\]

  2. AntiMatter
    • 3 years ago
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    all right, so if \[f(x)=e^{2x}\] differentiable on [0,3]?

  3. AntiMatter
    • 3 years ago
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    i'd say yes, we can graph the function and it is continuous and differentiable

  4. AntiMatter
    • 3 years ago
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    so, \[f'(x) = 2 * e^{2x}\]

  5. AntiMatter
    • 3 years ago
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    so evaluate f(3) and f(0)

  6. AntiMatter
    • 3 years ago
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    solve for the value c, such that \[f'(c) = 2*e^{2x} = (f(3) - f(0))/(3 - 0)\]

  7. AntiMatter
    • 3 years ago
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    \[f(3) = e^{6}, f(0) = 1\]

  8. AntiMatter
    • 3 years ago
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    so \[(e^6 - 1) / 3 = 2*e^{2*c}\] solve for c!

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