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Math Help - mean value theorem

  1. #1
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    mean value theorem

    Use the mean value theorem to establish the following inequalities.

    A) e^x > 1 + x for x>0
    B) (x-1)/x < ln x < x-1 for x >1
    e) sqrt(1+x) < 5 + (x-24)/10 for x>24
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  2. #2
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    Quote Originally Posted by luckyc1423 View Post
    Use the mean value theorem to establish the following inequalities.

    A) e^x > 1 + x for x>0
    Consider f(t)=e^t on the interval [0,x]

    Then, f is continous on [0,x] and differenciable on (0,x):

    (f(x)-f(0))/(x-0) = f'(c) for c in (0,x)

    Thus,
    (e^x-1)/x = e^c
    Thus,
    e^x-1= x*e^c for c>0 but then, e^c >1
    Thus,
    e^x - 1 > x
    Thus,
    e^x > 1+x
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  3. #3
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    Quote Originally Posted by luckyc1423 View Post
    B) (x-1)/x < ln x < x-1 for x >1
    Consider f(t) = ln t on the interval [1,x].

    Notice that f is continous on [1,x] and differenciable on (1,x).

    By Lagrange's Mean Value Theorem:

    (f(x)-f(1))/(x-1) = f'(c) for c in (1,x)

    Thus,

    ln(x)/(x-1) = 1/c for c in (1,x)

    Notice that
    1/x< 1/c < 1/1
    Thus,
    1/x < ln (x) / (x-1) < 1
    Multiply by denominator,
    (x-1)/x < ln x < (x-1)
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  4. #4
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    Quote Originally Posted by luckyc1423 View Post
    e) sqrt(1+x) < 5 + (x-24)/10 for x>24
    Consider,
    f(t) = sqrt(1+t) on interval [24,x].

    Thus, f is continous on [24,x] and differenciable on (24,x).

    Use the Mean Value Theorem:

    [sqrt(x+1) - sqrt(24+1)]/[x-24] = f'(c) = 1/2*sqrt(1+c) for c in (24,x).

    Note that, 1/2*sqrt(1+c) < 1/2*sqrt(24+1) = 1/10

    Thus,

    [sqrt(x+1) - 5]/[x-24] < 1/10

    Thus,

    sqrt(1+x) < 5 + (x-24)/10
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