Question-8410

Question Number 8410 by arinto27 last updated on 10/Oct/16

Answered by ridwan balatif last updated on 10/Oct/16

f(x)=5−2x, g(x)=x+6, h(x)=x−4 (fogoh)(x)=f(g(h(x))) =f(g(x−4)) =f(x+2) =−2x+1 (fogoh)(x)=−2x+1 y =−2x+1 y−1=−2x x =((1−y)/2) (fogoh)^(−1) (x)=((1−x)/2) (fogoh)^(−1) (k)=((1−k)/2) 2=((1−k)/2) 4=1−k⇒k=−3

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5}−\mathrm{2x},\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{6},\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{x}−\mathrm{4} \\ $$$$\left(\mathrm{fogoh}\right)\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{h}\left(\mathrm{x}\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}−\mathrm{4}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{f}\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2x}+\mathrm{1} \\ $$$$\left(\mathrm{fogoh}\right)\left(\mathrm{x}\right)=−\mathrm{2x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:\:\:\:=−\mathrm{2x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}−\mathrm{1}=−\mathrm{2x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:\:=\frac{\mathrm{1}−\mathrm{y}}{\mathrm{2}} \\ $$$$\left(\mathrm{fogoh}\right)^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{1}−\mathrm{x}}{\mathrm{2}} \\ $$$$\left(\mathrm{fogoh}\right)^{−\mathrm{1}} \left(\mathrm{k}\right)=\frac{\mathrm{1}−\mathrm{k}}{\mathrm{2}} \\ $$$$\mathrm{2}=\frac{\mathrm{1}−\mathrm{k}}{\mathrm{2}} \\ $$$$\mathrm{4}=\mathrm{1}−\mathrm{k}\Rightarrow\mathrm{k}=−\mathrm{3} \\ $$

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