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so as the title says Find all numbers $c$ that satisfies the conclusion of the Mean Value Theorem for the function $f(x) = x^3 −1$ on the interval $[−1,1]$.

i tried to solve it and it satisfy the conclusion and there should be a $C$, but when i solved it there is no $c$ because $x^2=-\frac13$ which is impossible and there no $c$ here, it ok?

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  • $\begingroup$ What does the $c$ in your problem mean? $\endgroup$
    – ViHdzP
    Dec 9 '19 at 19:18
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We want to find all numbers $c \in \langle -1,1\rangle$ such that $$f(1) - f(-1) = f'(c)(1-(-1))$$

or $$2 = 3c^2 \cdot 2$$

We get $$c^2 = \frac13 \implies c = \pm \frac1{\sqrt3}$$

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