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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