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

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

Answered by MJS last updated on 28/Jun/18

$$\mathrm{some}\mathrm{read}h\left(g^2\left(f\left(2\right)\right)\right),\mathrm{some}f\left(g^2\left(h\left(2\right)\right)\right)$$$$\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{case}\mathrm{we}\mathrm{have}$$$$h\left(x\right)=\frac{1}{2}\mathrm{for}\mathrm{all}x,\mathrm{so}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{to}\mathrm{care}$$$$\mathrm{about}\mathrm{the}\mathrm{rest}\mathrm{and}h\left(g^2\left(f\left(2\right)\right)\right)=\frac{1}{2}$$$$\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{case}\mathrm{we}\mathrm{have}$$$$h\left(2\right)=\frac{1}{2}$$$$g^2\left(\frac{1}{2}\right)={(\sqrt{\frac{1}{2}}+3)}^2={(\frac{\sqrt{2}}{2}+3)}^2=\frac{19}{2}+3\sqrt{2}$$$$f(\frac{19}{2}+3\sqrt{2})=20+6\sqrt{2}$$

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