If the function f(x) = x^3 - 6ax^2 + 5x satis | vedclass

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(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (1, 2)$ such that $f'(c) = \frac{f(2) - f(1)}{2 - 1}$.Given $f(x) = x^3 - 6

Vedclass · Accepted Answer

$\frac{35}{48}$

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(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (1, 2)$ such that $f'(c) = \frac{f(2) - f(1)}{2 - 1}$.Given $f(x) = x^3 - 6ax^2 + 5x$,we have:$f(2) = 2^3 - 6a(2^2) + 5(2) = 8 - 24a + 10 = 18 - 24a$.$f(1) = 1^3 - 6a(1^2) + 5(1) = 1 - 6a + 5 = 6 - 6a$.Thus,the slope of the chord is $\frac{f(2) - f(1)}{2 - 1} = (18 - 24a) - (6 - 6a) = 12 - 18a$.The derivative is $f'(x) = 3x^2 - 12ax + 5$.Since the tangent at $x = \frac{7}{4}$ is parallel to the chord,$f'(\frac{7}{4}) = 12 - 18a$.$3(\frac{7}{4})^2 - 12a(\frac{7}{4}) + 5 = 12 - 18a$.$3(\frac{49}{16}) - 21a + 5 = 12 - 18a$.$\frac{147}{16} + 5 - 12 = 21a - 18a$.$\frac{147 - 112}{16} = 3a$.$\frac{35}{16} = 3a$.$a = \frac{35}{48}$.

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