For the curve y = x^3 in the interval [-2, 2] | vedclass

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(C) Given the function $f(x) = x^3$ on the interval $[a, b] = [-2, 2]$.According to the Mean Value Theorem,there exists at least one point $c \in (-2,

Vedclass · Accepted Answer

$\pm \frac{2}{\sqrt{3}}$

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(C) Given the function $f(x) = x^3$ on the interval $[a, b] = [-2, 2]$.According to the Mean Value Theorem,there exists at least one point $c \in (-2, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.First,calculate the slope of the secant line:$f(2) = 2^3 = 8$$f(-2) = (-2)^3 = -8$Slope $= \frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - (-8)}{2 + 2} = \frac{16}{4} = 4$.Next,find the derivative of the function:$f'(x) = 3x^2$.Set the derivative equal to the slope of the secant line:$3c^2 = 4$$c^2 = \frac{4}{3}$$c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}}$.Thus,the abscissae are $\pm \frac{2}{\sqrt{3}}$.

For the curve $y = x^3$ in the interval $[-2, 2]$,find the abscissae of the points where the slope of the tangent is equal to the slope of the secant line passing through the endpoints of the interval,as per the Mean Value Theorem.

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