Question Number 74355 by Mr. K last updated on 22/Nov/19
$${let}\:{f}\left({x}\right),\:{g}\left({x}\right)\:{and}\:{h}\left({x}\right)\:{be}\:{functions} \\ $$$$\mathbb{R}\rightarrow\mathbb{R},\:{given}\:{by} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} ,\:{if}\:{x}\geqslant\mathrm{0}\:{and}\:{x}+\mathrm{1}\:{if}\:{x}<\mathrm{0} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{4},\:{if}\:{x}\geqslant\mathrm{2}\:{and}\:\frac{\mathrm{1}}{\mathrm{2}−{x}}\:{if}\:{x}<\mathrm{2} \\ $$$${h}\left({x}\right)=\mathrm{3}^{−{x}} ,\:{if}\:{x}\leqslant\mathrm{0}\:{and}\:\mathrm{3}^{{x}} \:{if}\:{x}\geqslant\mathrm{0} \\ $$$${Calculate}\:\frac{{f}\left(\mathrm{2}\right)+{f}\left({g}\left(\mathrm{2}\right)\right)}{{f}\left({g}\left({h}\left(−\mathrm{1}\right)\right)\right)}. \\ $$
Answered by MJS last updated on 23/Nov/19
$$\mathrm{easy} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{4} \\ $$$${g}\left(\mathrm{2}\right)=\mathrm{0}\:{f}\left({g}\left(\mathrm{2}\right)\right)={f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${h}\left(−\mathrm{1}\right)=\mathrm{3}\:{g}\left({h}\left(−\mathrm{1}\right)\right)={g}\left(\mathrm{3}\right)=\mathrm{5}\:{f}\left({g}\left({h}\left(−\mathrm{1}\right)\right)\right)={f}\left(\mathrm{5}\right)=\mathrm{25} \\ $$$$\frac{{f}\left(\mathrm{2}\right)+{f}\left({g}\left(\mathrm{2}\right)\right)}{{f}\left({g}\left({h}\left(−\mathrm{1}\right)\right)\right)}=\frac{\mathrm{4}}{\mathrm{25}} \\ $$