On the curve y=x^3,the point at which the tan | vedclass

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(C) Let $f(x) = x^3$. The slope of the chord joining the points $(-1, -1)$ and $(2, 8)$ is given by $m = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - (-1

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$(1, 1)$

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(C) Let $f(x) = x^3$. The slope of the chord joining the points $(-1, -1)$ and $(2, 8)$ is given by $m = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - (-1)}{3} = \frac{9}{3} = 3$. Since the tangent line is parallel to this chord,the derivative $f'(x)$ at the point must be equal to the slope of the chord. $f'(x) = \frac{d}{dx}(x^3) = 3x^2$. Setting $f'(x) = 3$,we get $3x^2 = 3$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$. For $x = 1$,$y = (1)^3 = 1$,giving the point $(1, 1)$. For $x = -1$,$y = (-1)^3 = -1$,giving the point $(-1, -1)$. Since $(-1, -1)$ is one of the endpoints of the chord,the point on the curve between the given points is $(1, 1)$.

On the curve $y=x^3$,the point at which the tangent line is parallel to the chord joining the points $(-1, -1)$ and $(2, 8)$ is

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