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Thread: rule of differentiation: mean value theorem

  1. #1
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    rule of differentiation: mean value theorem

    hi, guys. i'm having trouble solving this problem. i have a feeling i am making it more difficult than it really is. I really need help because the hw is due today! please help!

    show that (1+x)^(1/2) < 1 + (1/2)x if x > 0
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  2. #2
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    anyone????
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  3. #3
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    Hi,

    The first thing I'd do is get rid of the square root by squaring both sides of the equation. then cancel out some x's and units, and you should be just about there with the proof.
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  4. #4
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    yea, i get how to do it algebraically, but it says to use the mean value theorem. that's what has me confused.
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  5. #5
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    Suppose that x > 0. Because f(x) = \sqrt {1 + x} has a derivative, by the mean value theorem
    \left( {\exists c \in \left( {0,x} \right)} \right)\left[ {\frac{1}{{2\sqrt {1 + c} }} = \frac{{\sqrt {1 + x}  - 1}}{x}} \right].
    But we know that \frac{1}{{2\sqrt {1 + c} }} < \frac{1}{2}.
    Now you put all that together to finish it all off.
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  6. #6
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    wow, ok, i see. thank you soooo much!
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