# The points on the graph $y = {x^3} - 3x$ at which the tangent is parallel to x-axis are:

A).(2,2) and (1,-2)

B).(-1,2) and (-2,-2)

C).(2,2) and (-1,2)

D).(-2,-2) and (2,2)

E).(1,-2) and (-1,2)

Answer

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Hint: Any line which is parallel to x-axis will have the slope equal to zero.and differentiate the given curve to get the solution.

Given, curve is $y = {x^3} - 3x \to (1)$

On differentiating the above curve equation with respect to $x,$we get

$\dfrac{{dy}}{{dx}} = 3{x^2} - 3$

Since, tangent is parallel to $x$-axis. (Given)

Therefore, $\dfrac{{dy}}{{dx}} = 0$

$ \Rightarrow 3{x^2} - 3 = 0$

$ \Rightarrow {x^2} = 1$

$ \Rightarrow x = \pm 1$

From equation (1), we have

When $x = 1,y = {1^3} - 3(1) = - 2$

When $x = - 1,y = {( - 1)^3} - 3( - 1) = 2$

Therefore, required points are (1, -2) and (-1, 2).

Note: A line parallel to the x-axis will have slope m=0. So you need to take the first derivative, and set it equal to zero to solve for the x values at which the slope of the tangent to your curve is zero.

Given, curve is $y = {x^3} - 3x \to (1)$

On differentiating the above curve equation with respect to $x,$we get

$\dfrac{{dy}}{{dx}} = 3{x^2} - 3$

Since, tangent is parallel to $x$-axis. (Given)

Therefore, $\dfrac{{dy}}{{dx}} = 0$

$ \Rightarrow 3{x^2} - 3 = 0$

$ \Rightarrow {x^2} = 1$

$ \Rightarrow x = \pm 1$

From equation (1), we have

When $x = 1,y = {1^3} - 3(1) = - 2$

When $x = - 1,y = {( - 1)^3} - 3( - 1) = 2$

Therefore, required points are (1, -2) and (-1, 2).

Note: A line parallel to the x-axis will have slope m=0. So you need to take the first derivative, and set it equal to zero to solve for the x values at which the slope of the tangent to your curve is zero.

Last updated date: 21st Sep 2023

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