Question Number 56852 by peter frank last updated on 25/Mar/19
Answered by MJS last updated on 25/Mar/19
$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:{f}\circ{g}\:\mathrm{means}\:{f}\left({g}\right)\:\mathrm{or}\:{g}\left({f}\right) \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{only}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{thinking}\:\mathrm{logically} \\ $$$${f}\left({g}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{3};\:−\mathrm{1}\leqslant{x}<−\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{4}{x}^{\mathrm{2}} ;\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{2};\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\\{\mathrm{1};\:{x}>\mathrm{1}}\end{cases} \\ $$$${g}\left({f}\left({x}\right)\right)=\begin{cases}{\mathrm{1};\:{x}<−\mathrm{1}}\\{\mathrm{2}{x}^{\mathrm{2}} ;\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}}\\{\mathrm{1};\:{x}>\mathrm{1}}\end{cases} \\ $$
Commented by rahul 19 last updated on 25/Mar/19
fog means f(g) and gof means g(f).
Commented by MJS last updated on 25/Mar/19
$$\mathrm{interestingly}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{standardized} \\ $$