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Thread: Intermediate Value Theorem

  1. #1
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    Intermediate Value Theorem

    Use the Intermediate Value Theorem to show that the polynomial has a zero in the given interval.

    f(x) = 3x^4 + 4x^3 - 8x - 2; [1, 2]

    Also, I noticed that there is no x^2 term. Does this change the process in terms of solving the question?



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  2. #2
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    Quote Originally Posted by magentarita View Post
    Use the Intermediate Value Theorem to show that the polynomial has a zero in the given interval.

    f(x) = 3x^4 + 4x^3 - 8x - 2; [1, 2]

    Also, I noticed that there is no x^2 term. Does this change the process in terms of solving the question?


    Let $\displaystyle f(x)=3x^4+4x^3-8x-2$. Since f is a polynomial, it is continuous on $\displaystyle [1, 2]$.

    We have $\displaystyle f(1)=3(1)^4+4(1)^3-8x-2=-3 < 0 \ \ and \ \ f(2)=3(2)^4+4(2)^3-8x-2=62 > 0$, so that $\displaystyle f(1)<0<f(2)$.

    So by the intermediate-value theorem there exists $\displaystyle x_1$ in $\displaystyle (1, 2)$ such that $\displaystyle f(x_1)=0$.

    That is, the equation $\displaystyle 3x^4+4x^3-8x-2=0$ has a solution in the interval $\displaystyle (1, 2)$.
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  3. #3
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    ok....

    Quote Originally Posted by masters View Post
    Let $\displaystyle f(x)=3x^4+4x^3-8x-2$. Since f is a polynomial, it is continuous on $\displaystyle [1, 2]$.

    We have $\displaystyle f(1)=3(1)^4+4(1)^3-8x-2=-3 < 0 \ \ and \ \ f(2)=3(2)^4+4(2)^3-8x-2=62 > 0$, so that $\displaystyle f(1)<0<f(2)$.

    So by the intermediate-value theorem there exists $\displaystyle x_1$ in $\displaystyle (1, 2)$ such that $\displaystyle f(x_1)=0$.

    That is, the equation $\displaystyle 3x^4+4x^3-8x-2=0$ has a solution in the interval $\displaystyle (1, 2)$.
    Thank you very much.
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