The abscissa of the points of the curve y = x | vedclass

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

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

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(A) Given the curve $f(x) = x^3$ on the interval $[-2, 2]$.According to the Mean Value Theorem,there exists at least one point $c \in (-2, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = -2$ and $b = 2$.$f(2) = 2^3 = 8$ and $f(-2) = (-2)^3 = -8$.The slope of the secant line is $\frac{f(2) - f(-2)}{2 - (-2)} = \frac{8 - (-8)}{4} = \frac{16}{4} = 4$.The derivative is $f'(x) = 3x^2$.Setting $f'(c) = 4$,we get $3c^2 = 4$.$c^2 = \frac{4}{3}$,which implies $c = \pm \frac{2}{\sqrt{3}}$.

The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

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