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Question Number 106044 by ZiYangLee last updated on 02/Aug/20

Given f(x)=2x+m, f^(2 ) (x)=px+6 Find the value of m and p.

$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}+\mathrm{m},\:\mathrm{f}^{\mathrm{2}\:} \left(\mathrm{x}\right)=\mathrm{px}+\mathrm{6} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{m}\:\mathrm{and}\:\mathrm{p}. \\ $$

Answered by 1549442205PVT last updated on 02/Aug/20

f^2 (x)=4x^2 +4mx+m^2 . f^2 (x)=px+6⇔4x^2 +4mx+m^2 =px+6 ⇒p=ax+b⇒4x^2 +4mx+m^2 =(ax+b)x+6 ⇔4x^2 +4mx+m^2 =ax^2 +bx+6 ⇒ { ((a=4)),((4m=b)),((m^2 =6)) :} ⇔ { ((m=±(√6))),((b=±4(√6))),((a=4)) :} Thus, p=4x±4(√6) ,m=±(√6)

$$\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{4x}^{\mathrm{2}} +\mathrm{4mx}+\mathrm{m}^{\mathrm{2}} . \\ $$$$\mathrm{f}\:^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{px}+\mathrm{6}\Leftrightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{4mx}+\mathrm{m}^{\mathrm{2}} =\mathrm{px}+\mathrm{6} \\ $$$$\Rightarrow\mathrm{p}=\mathrm{ax}+\mathrm{b}\Rightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{4mx}+\mathrm{m}^{\mathrm{2}} =\left(\mathrm{ax}+\mathrm{b}\right)\mathrm{x}+\mathrm{6} \\ $$$$\Leftrightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{4mx}+\mathrm{m}^{\mathrm{2}} =\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{6} \\ $$$$\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{4}}\\{\mathrm{4m}=\mathrm{b}}\\{\mathrm{m}^{\mathrm{2}} =\mathrm{6}}\end{cases}\:\:\Leftrightarrow\begin{cases}{\mathrm{m}=\pm\sqrt{\mathrm{6}}}\\{\mathrm{b}=\pm\mathrm{4}\sqrt{\mathrm{6}}}\\{\mathrm{a}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{Thus},\:\mathrm{p}=\mathrm{4x}\pm\mathrm{4}\sqrt{\mathrm{6}}\:,\mathrm{m}=\pm\sqrt{\mathrm{6}} \\ $$

Commented by ZiYangLee last updated on 03/Aug/20

thanks <3

$$\mathrm{thanks}\:<\mathrm{3} \\ $$

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