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Find the equation of tangent to the curv...

Find the equation of tangent to the curve `y= x^(3)-x` , at the point at which slope of tangent is equal to zero.

Text Solution

Verified by Experts

Slope of tangent to the curve is given by
`(dy)/(dx)=3x^(2)-1`
Given that slope of tangent =0
So, `(dy)/(dx)=0`
`implies 3x^(2)-1=0`
`implies x= pm (1)/(sqrt(3))`
For , `x =(1)/(sqrt(3)) , y=-(2)/(3sqrt(3))`
Equation of tangent is `(y+(2)/(3 sqrt(3)))=0`
For , ` x=-(1)/(sqrt(3)) , y=(2)/(3sqrt(3))`
Equation of tangent is `y-(2)/(3sqrt(3))=0`
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