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Find the number c that satisfies the conclusion of Rolle's Theorem. f(x) = x3 - x2 - 2x + 8 [0, 2] ...c=?
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If a real-valued function ƒ is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and ƒ(a) = ƒ(b), then there exists a c in the open interval (a, b) such that\[f'(c)=0\]That's Rolle's Theorem. All you need to do is take the first derivative of what you've got, take the function of c, set it to zero and solve. So,\[f'(x)=3x^2-2x-2 \rightarrow f'(c)=3c^2-2c-2:=0\]You need to find c such that\[3c^2-2c-2=0\]
When you solve for c, you'll end up with two solutions,\[c=\frac{1 \pm \sqrt{7}}{3}\]Only one of them lies in the interval [0,2], namely,\[\frac{1+\sqrt{7}}{3}\]The other one, you have to reject.
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