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Given f9x0=x2=3x-5 on {-1,3} show f(x)= 10 without solving for x

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  1. The question as it is is not very clear. I will restate it the way I understand it. If the restatement is incorrect, please give the corrected version.

    "Given f(x)=x²+3x-5 on [-1,3] show that f(x)= 10 exists without solving for x."

    We first note that f(x) is a polynomial, and is therefore continuous and differentiable throughout its domain [-1,3].

    Then we evaluate f(-1)=-7 and f(3)=13.
    Thus f(-1)<10<f(3). By the intermediate value theorem, we conclude that c exists for which f(c)=10, or more precisely,
    ∃ -1<c<3 such that f(-1)<f(c)<f(3).

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