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

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(D) Given curve is $y=x^3$.The slope of the tangent to the curve at any point $(x_1, y_1)$ is given by $\frac{dy}{dx}$.Differentiating $y=x^3$ with re

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$(0,0)$

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(D) Given curve is $y=x^3$.The slope of the tangent to the curve at any point $(x_1, y_1)$ is given by $\frac{dy}{dx}$.Differentiating $y=x^3$ with respect to $x$,we get $\frac{dy}{dx} = 3x^2$.The slope of the tangent at $(x_1, y_1)$ is $m_T = 3x_1^2$.Since the tangent is parallel to the $X$-axis,its slope must be equal to the slope of the $X$-axis,which is $0$.Therefore,$m_T = 0$,which implies $3x_1^2 = 0$,so $x_1 = 0$.Substituting $x_1 = 0$ into the curve equation $y=x^3$,we get $y_1 = (0)^3 = 0$.Thus,the required point is $(0,0)$.

The point on the curve $y=x^3$,at which the tangent to the curve is parallel to the $X$-axis,is

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