Posted by Mark on Monday, June 11, 2012 at 9:38pm.
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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