Consider the function f(x) = 8x^2 - 7x + 5 on | vedclass

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(D) The Mean Value Theorem states that there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 8x^2 - 7x

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(D) The Mean Value Theorem states that there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 8x^2 - 7x + 5$ on $[-6, 6]$,we have $a = -6$ and $b = 6$.First,calculate $f(6) = 8(6)^2 - 7(6) + 5 = 8(36) - 42 + 5 = 288 - 42 + 5 = 251$.Next,calculate $f(-6) = 8(-6)^2 - 7(-6) + 5 = 8(36) + 42 + 5 = 288 + 42 + 5 = 335$.The derivative is $f'(x) = 16x - 7$,so $f'(c) = 16c - 7$.Applying the formula: $16c - 7 = \frac{f(6) - f(-6)}{6 - (-6)} = \frac{251 - 335}{12} = \frac{-84}{12} = -7$.Thus,$16c - 7 = -7$,which implies $16c = 0$,so $c = 0$.

Consider the function $f(x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the Mean Value Theorem is:

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