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Question Number 161130 by blackmamba last updated on 12/Dec/21

G i v e n P (x) i s p o l y n o m i a l s u c h t h a t

P (3 x) = P^{'} (x) . P^{″} (x) . F i n d t h e t a n g e n t

o f c u r v e y = P (x) p a r a l l e l t o t h e l i n e

y = 4 x - 2 .

Answered by FongXD last updated on 12/Dec/21

Given : P (3 x) = P^{'} (x) \cdot P^{″} (x)

Degree of L . H . S = n

Degree of R . H . S = (n - 1) + (n - 2) = 2 n - 3

L . H . S = R . H . S, \Leftrightarrow n = 2 n - 3, \Rightarrow n = 3

therefore, P (x) = {ax}^{3} + {bx}^{2} + cx + d

then P (3 x) = {27 ax}^{3} + {9 bx}^{2} + 3 cx + d

∙ P^{'} (x) = {3 ax}^{2} + 2 bx + c

∙ P^{″} (x) = 6 ax + 2 b

\Rightarrow P^{'} (x) \cdot P^{″} (x) = {18 a}^{2} x^{3} + {18 abx}^{2} + ({4 b}^{2} + 6 ac) x + 2 bc

P (3 x) = P^{'} (x) \cdot P^{″} (x)

then {\begin{cases} 27 a = {18 a}^{2} \\ {9 b}^{2} = 18 ab \\ 3 c = {4 b}^{2} + 6 ac \\ d = 2 bc \end{cases}, \Rightarrow {\begin{cases} a = \frac{3}{2} \\ b = 3 \\ 3 c = 36 + 9 c, \Rightarrow c = - 6 \\ d = - 36 \end{cases}

then P (x) = \frac{3}{2} x^{3} + {3 x}^{2} - 6 x - 36

and P^{'} (x) = \frac{9}{2} x^{2} + 6 x - 6

Given that tangent line is parallel to the line y = 4 x - 2

\Rightarrow P^{'} (x_{0}) = 4, \Leftrightarrow \frac{9}{2} x_{0}^{2} + {6 x}_{0} - 6 = 4

\Leftrightarrow {9 x}_{0}^{2} + {12 x}_{0} - 20 = 0

\Rightarrow x_{0} = \frac{- 6 \pm \sqrt{36 + 180}}{9} = \frac{- 2 \pm 2 \sqrt{6}}{3}

\Rightarrow P (x_{0}) = \frac{3}{2} {(\frac{- 2 \pm 2 \sqrt{6}}{3})}^{3} + 3 {(\frac{- 2 \pm 2 \sqrt{6}}{3})}^{2} - 6 (\frac{- 2 \pm 2 \sqrt{6}}{3}) - 36

\Rightarrow {\begin{cases} P (x_{0}) = \frac{- 280 - 24 \sqrt{6}}{9}, if x_{0} = \frac{- 2 + 2 \sqrt{6}}{3} \\ P (x_{0}) = \frac{- 280 + 24 \sqrt{6}}{9}, if x_{0} = \frac{- 2 - 2 \sqrt{6}}{3} \end{cases}

Tangent line : y = P^{'} (x_{0}) (x - x_{0}) + P (x_{0})

\Rightarrow {\begin{cases} y = 4 (x - \frac{- 2 + 2 \sqrt{6}}{3}) + \frac{- 280 - 24 \sqrt{6}}{9} \\ y = 4 (x - \frac{- 2 - 2 \sqrt{6}}{3}) + \frac{- 280 + 24 \sqrt{6}}{9} \end{cases}

\Rightarrow {\begin{cases} y = 4 x - \frac{256 + 48 \sqrt{6}}{9} \\ y = 4 x - \frac{256 - 48 \sqrt{6}}{9} \end{cases}

Commented by cortano last updated on 12/Dec/21

9 b = 18 a b \to b = 0

P (x) = \frac{3}{2} x^{3} \to {\begin{cases} P (3 x) = \frac{81}{2} x^{3} \\ P^{'} (x) = \frac{9}{2} x^{2} \\ P^{″} (x) = 9 x \end{cases}

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