Find the values of $c$ that satisfy the Rolle's theorem for integrals on $[- 1, 3]$ .
$f (x) = x^{2} - 2 x - 8$

c = 3

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c = 1

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c = 0

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c = 2

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Solution

The correct option is B $c = 1$
Given : $f (x) = x^{2} - 2 x - 8 (- 1, 3)$
Rolles's theorem : $f^{'} (c) = \frac{f (b) - f (a)}{(b - a)}$
$(2 c - 2) = \frac{f (3) - f (- 1)}{(3 - (- 1))} = \frac{(9 - 6 - 8) - (1 + 2 - 8)}{(4)} = 0 2 c - 2 = 0 c = 1$
Hence the correct answer is $c = 1$

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