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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 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangent}\left(\mathrm{T}_{\mathrm{0}} \right)\:\mathrm{to}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{x} \\ $$$$\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =\mathrm{2}.\:\mathrm{Find}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{T}_{\mathrm{1}} \:\mathrm{at}\:\mathrm{x}_{\mathrm{1}} \\ $$$$\mathrm{be}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{T}_{\mathrm{0}} . \\ $$

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)=3x^2 −2x−1⇒f ′(2)=12−5=7 the eq of tangent line at x=2 can be written as 7x−y=7.(2)−1(2) 7x−y=12

$$\mathrm{solution}\::\:\mathrm{y}_{\mathrm{0}} =\mathrm{8}−\mathrm{4}−\mathrm{2}\:=\:\mathrm{2} \\ $$$$\mathrm{gradient}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{f}\:'\left(\mathrm{2}\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{1}\Rightarrow\mathrm{f}\:'\left(\mathrm{2}\right)=\mathrm{12}−\mathrm{5}=\mathrm{7} \\ $$$$\mathrm{the}\:\mathrm{eq}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{x}=\mathrm{2} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\mathrm{7x}−\mathrm{y}=\mathrm{7}.\left(\mathrm{2}\right)−\mathrm{1}\left(\mathrm{2}\right) \\ $$$$\mathrm{7x}−\mathrm{y}=\mathrm{12} \\ $$

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

slope at x_1 = f ′(x_1 ) = 7 [ since parallel to x_0 ] 3x_1 ^2 −2x_1 −8=0 (3x_1 +4 )(x_1 −2) = 0 x_1 = 2 = x_0 ( not satisfy to condition) so x_1 =−(4/3)

$$\mathrm{slope}\:\mathrm{at}\:\mathrm{x}_{\mathrm{1}} \:=\:\mathrm{f}\:'\left(\mathrm{x}_{\mathrm{1}} \right)\:=\:\mathrm{7}\:\left[\:\mathrm{since}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{x}_{\mathrm{0}} \right] \\ $$$$\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2x}_{\mathrm{1}} −\mathrm{8}=\mathrm{0} \\ $$$$\left(\mathrm{3x}_{\mathrm{1}} \:+\mathrm{4}\:\right)\left(\mathrm{x}_{\mathrm{1}} −\mathrm{2}\right)\:=\:\mathrm{0}\: \\ $$$$\mathrm{x}_{\mathrm{1}} =\:\mathrm{2}\:=\:\mathrm{x}_{\mathrm{0}} \:\left(\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{to}\:\mathrm{condition}\right) \\ $$$$\mathrm{so}\:\mathrm{x}_{\mathrm{1}} =−\frac{\mathrm{4}}{\mathrm{3}}\: \\ $$

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