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f-f-n-2n-f-n-

Question Number 32743 by naka3546 last updated on 01/Apr/18 $${f}\:\left({f}\:\left({n}\right)\right)\:\:=\:\:\mathrm{2}{n} \\ $$$${f}\:\left({n}\right)\:\:=\:\:? \\ $$ Commented by abdo imad last updated on 03/Apr/18 $${if}\:{f}\:{is}\:{a}\:{polynomial}\:{fof}\left({x}\right)=\mathrm{2}{x}\:\Rightarrow{f}^{,} \left({f}\left({x}\right)\right){f}^{'} \left({x}\right)=\mathrm{2}\:\Rightarrow{y}^{,}…

Question-32710

Question Number 32710 by mondodotto@gmail.com last updated on 31/Mar/18 Answered by MJS last updated on 01/Apr/18 $${f}\left({x}\right)={a}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{3} \\ $$$${a}×\mathrm{1}×\left(−\mathrm{4}\right)=\mathrm{3} \\ $$$${a}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}}…

Question-32709

Question Number 32709 by mondodotto@gmail.com last updated on 31/Mar/18 Answered by MJS last updated on 31/Mar/18 $$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}=\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{p} \\ $$$$\mathrm{log}_{\mathrm{3}} \:\mathrm{2}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}=\mathrm{p}^{−\mathrm{1}} \\ $$$$\mathrm{2}^{{p}^{{x}} } =\mathrm{3}^{{p}^{−{x}}…

Question-32666

Question Number 32666 by naka3546 last updated on 30/Mar/18 Commented by abdo imad last updated on 30/Mar/18 $${let}\:{put}\:{A}_{{n}} =\:\frac{\left({n}+\mathrm{6}\right)^{\frac{{n}+\mathrm{6}}{{n}}} \:−{n}^{\frac{{n}+\mathrm{6}}{{n}}} }{\left({n}+\mathrm{3}\right)^{\frac{{n}+\mathrm{3}}{{n}}} \:−\:{n}^{\frac{{n}+\mathrm{3}}{{n}}} } \\ $$$${A}_{{n}}…

Given-the-function-f-x-x-1-3x-and-g-x-x-1-Find-a-fg-b-f-g-c-gf-1-x-

Question Number 32640 by Rio Mike last updated on 30/Mar/18 $${Given}\:{the}\:{function}\:{f}:{x}\rightarrow\:\frac{{x}\:+\mathrm{1}}{\mathrm{3}{x}} \\ $$$${and}\:{g}\::\:{x}\:\rightarrow\:{x}−\mathrm{1}.\mathrm{Find}\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{fg} \\ $$$$\left.\mathrm{b}\right)\mathrm{f}°\mathrm{g} \\ $$$$\left.\mathrm{c}\right)\:{gf}^{−\mathrm{1}} \left({x}\right) \\ $$ Answered by Joel578…

Question-32635

Question Number 32635 by mondodotto@gmail.com last updated on 29/Mar/18 Commented by Joel578 last updated on 30/Mar/18 $$\mathrm{Use}\:\mathrm{substitution} \\ $$$${u}\:=\:{e}^{{e}^{\iddots^{{x}} } } \:\:\left(\mathrm{8}\:{e}\:\mathrm{as}\:\mathrm{powers}\right) \\ $$$$\mathrm{and}\:\mathrm{gives}\:\mathrm{you} \\…