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Question Number 38692 by Zuarkton last updated on 28/Jun/18

If f(x)=2x+1 g(x)=(√x)+3 h(x)=(1/2) then hog^2 of (2)=?

$${If}\:{f}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${g}\left({x}\right)=\sqrt{{x}}+\mathrm{3} \\ $$$${h}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${then}\:{hog}^{\mathrm{2}} \:{of}\:\left(\mathrm{2}\right)=? \\ $$

Answered by MJS last updated on 28/Jun/18

some read h(g^2 (f(2))), some f(g^2 (h(2))) in the first case we have h(x)=(1/2) for all x, so we don′t have to care about the rest and h(g^2 (f(2)))=(1/2) in the second case we have h(2)=(1/2) g^2 ((1/2))=((√(1/2))+3)^2 =(((√2)/2)+3)^2 =((19)/2)+3(√2) f(((19)/2)+3(√2))=20+6(√2)

$$\mathrm{some}\:\mathrm{read}\:{h}\left({g}^{\mathrm{2}} \left({f}\left(\mathrm{2}\right)\right)\right),\:\mathrm{some}\:{f}\left({g}^{\mathrm{2}} \left({h}\left(\mathrm{2}\right)\right)\right) \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have} \\ $$$${h}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{for}\:\mathrm{all}\:{x},\:\mathrm{so}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{care} \\ $$$$\mathrm{about}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{and}\:{h}\left({g}^{\mathrm{2}} \left({f}\left(\mathrm{2}\right)\right)\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have} \\ $$$${h}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${g}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}+\mathrm{3}\right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{3}\right)^{\mathrm{2}} =\frac{\mathrm{19}}{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${f}\left(\frac{\mathrm{19}}{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{2}}\right)=\mathrm{20}+\mathrm{6}\sqrt{\mathrm{2}} \\ $$

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