Find-the-equation-of-the-tangent-T-0-to-y-x-3-x-2-x-at-x-0-2-Find-x-1-such-that-the-tangent-T-1-at-x-1-be-parallel-to-T-0-

Question Number 95636 by Ar Brandon last updated on 26/May/20

Find the equation of the tangent (T_{0}) to y = x^{3} - x^{2} - x

at x_{0} = 2 . Find x_{1} such that the tangent T_{1} at x_{1}

be parallel to T_{0} .

Answered by john santu last updated on 26/May/20

solution : y_{0} = 8 - 4 - 2 = 2

gradient of tangent line f^{'} (2)

f^{'} (x) = {3 x}^{2} - 2 x - 1 \Rightarrow f^{'} (2) = 12 - 5 = 7

the eq of tangent line at x = 2

can be written as 7 x - y = 7 . (2) - 1 (2)

7 x - y = 12

Answered by john santu last updated on 26/May/20

slope at x_{1} = f^{'} (x_{1}) = 7 [since parallel to x_{0}]

{3 x}_{1}^{2} - {2 x}_{1} - 8 = 0

({3 x}_{1} + 4) (x_{1} - 2) = 0

x_{1} = 2 = x_{0} (not satisfy to condition)

so x_{1} = - \frac{4}{3}