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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Verify Mean Value Theorem for the function f(x)=x^2+lnx-1 on the interval [1,2]
    by finding (all) the appropriate point(s) c where the derivative equals the slope of the
    secant line between the end points of the interval.

    I think I know how to do this. But I have got to solve it without a calculator apparently!!

    slope of secant line is ln2+3
    f'(x)=2x+\frac{1}{x}
    let m = ln2 + 3
    2x+\frac{1}{x}=m
    2x^2+1=mx
    2x^2-mx+1
    \frac{m\pm\sqrt{(m^2-8)}}{4}
    now substitute:
    \frac{ln2 + 3\pm\sqrt{((ln2 + 3)^2-8)}}{4}

    now I'm stuck here. How do I solve this without a calculator?

    Thanks!
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  2. #2
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    Quote Originally Posted by Vamz View Post
    Verify Mean Value Theorem for the function f(x)=x^2+lnx-1 on the interval [1,2]
    by finding (all) the appropriate point(s) c where the derivative equals the slope of the
    secant line between the end points of the interval.

    I think I know how to do this. But I have got to solve it without a calculator apparently!!

    slope of secant line is ln2+3
    f'(x)=2x+\frac{1}{x}
    let m = ln2 + 3
    2x+\frac{1}{x}=m
    2x^2+1=mx
    2x^2-mx+1
    \frac{m\pm\sqrt{(m^2-8)}}{4}
    now substitute:
    \frac{ln2 + 3\pm\sqrt{((ln2 + 3)^2-8)}}{4}

    now I'm stuck here. How do I solve this without a calculator?

    Thanks!
    the value of x in the interval [1,2] that satisfies the conclusion of the MVT is x = \frac{\ln{2}+3 + \sqrt{(\ln{2}+3)^2 - 8}}{4} because \frac{\ln{2}+3}{4} < 1
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