Application of Derivatives

Find the points on the curve y = x 3 – 2x 2 – x at which the tangent lines are parallel to the line y = 3x – 2. - Wired Faculty

Question

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Find the points on the curve y = x³ – 2x² – x at which the tangent lines are parallel to the line y = 3x – 2.

Solution

The equation of curve is $straight y equals straight x cubed minus 2 straight x squared minus straight x$ ...(1)
$therefore space space space space dy over dx space equals space 3 straight x squared minus 4 straight x minus 1 At space left parenthesis straight x comma space straight y right parenthesis comma space space slope space of space tangent space equals 3 straight x squared minus 4 straight x minus 1 therefore space space space space space straight m subscript 1 space equals space 3 straight x squared minus 4 straight x minus 1 comma space space space where space straight m subscript 1 space is space slope space of space tangent Let space space straight m subscript 2 space be space slope space of space line space straight y equals space space 3 straight x minus 2 therefore space space space space straight m subscript 2 space equals space 3 because space space tangent space is space parallel space to space line space left parenthesis 2 right parenthesis therefore space space space space space straight m subscript 1 space equals space straight m subscript 2 space space space space rightwards double arrow space space space space space 3 straight x squared minus 4 straight x minus 1 space space space space space rightwards double arrow space space space space space 3 straight x squared minus 4 straight x minus 4 space equals space 0 rightwards double arrow space space space space space straight x space equals space fraction numerator 4 plus-or-minus square root of 16 plus 48 end root over denominator 6 end fraction space equals space fraction numerator 4 plus-or-minus 8 over denominator 6 end fraction space equals space 2 comma space space space space minus 2 over 3 When space straight x space equals space 2 comma space space space from space left parenthesis 1 right parenthesis comma space space space straight y space equals space 8 minus 8 minus 2 space equals space minus 2 When space straight x space equals space minus 2 over 3 comma space space from space left parenthesis 1 right parenthesis comma space space space straight y space equals space minus 8 over 27 minus 8 over 9 plus 2 over 3 space equals space fraction numerator negative 8 minus 24 plus 18 over denominator 27 end fraction space equals space minus 14 over 27 therefore space space space space space required space points space are space left parenthesis 2 comma space minus 2 right parenthesis comma space space space open parentheses negative 2 over 3 comma space space minus 14 over 27 close parentheses$

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