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Thread: Mean Value/Rolle's Theorem

  1. #1
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    Mean Value/Rolle's Theorem

    Consider the function $\displaystyle f(x)=6x^4-7x+1$ on the interval $\displaystyle [1/2,2].$ $\displaystyle f$ has at least one zero in $\displaystyle [1/2,2]$ (actually f(1)=0) so show that $\displaystyle f$ has *exactly* on zero in the indicated interval. Hint:Use the Mean Value Theorem/Rolle's Theorem to prove the result.
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  2. #2
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    Quote Originally Posted by Zocken View Post
    Consider the function $\displaystyle f(x)=6x^4-7x+1$ on the interval $\displaystyle [1/2,2].$ $\displaystyle f$ has at least one zero in $\displaystyle [1/2,2]$ (actually f(1)=0) so show that $\displaystyle f$ has *exactly* on zero in the indicated interval. Hint:Use the Mean Value Theorem/Rolle's Theorem to prove the result.
    if $\displaystyle f$ had more than one zero in [1/2, 2], then by Rolle's theorem $\displaystyle f'(x)=0$ would have at least one root in [1/2, 2]. but the (real) root of $\displaystyle f'(x)=24x^3-7=0$ is $\displaystyle x=\sqrt[3]{7/24} \notin [1/2 , 2].$
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  3. #3
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    wow makes it so simple. thanks a bunch!
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