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

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

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

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

Commented by ZiYangLee last updated on 03/Aug/20

$$\mathrm{thanks}<3$$

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