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Category: Relation and Functions

f-R-R-g-R-R-f-x-y-f-x-g-y-g-x-f-y-g-x-y-f-x-f-y-g-x-g-y-f-x-g-x-

Question Number 1979 by prakash jain last updated on 27/Oct/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({y}\right)+{g}\left({x}\right){f}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$ Answered by…

f-R-R-g-R-R-f-x-y-f-x-g-x-f-y-g-y-g-x-y-f-x-f-y-g-x-g-y-f-x-2-g-x-2-f-x-g-y-g-x-f-y-f-x-g-x-

Question Number 1976 by 123456 last updated on 27/Oct/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right) \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} =? \\ $$$$\left[{f}\left({x}\right)+{g}\left({y}\right)\right]\left[{g}\left({x}\right)+{f}\left({y}\right)\right]=?? \\ $$$${f}\left({x}\right)=??? \\…

f-0-R-a-N-R-a-n-1-f-a-n-a-n-f-x-f-y-x-y-0-does-a-n-a-m-n-m-0-

Question Number 1963 by 123456 last updated on 26/Oct/15 $${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R},{a}:\mathbb{N}\rightarrow\mathbb{R} \\ $$$${a}_{{n}+\mathrm{1}} ={f}\left({a}_{{n}} \right)−{a}_{{n}} \\ $$$${f}\left({x}\right)\geqslant{f}\left({y}\right),\forall{x}\geqslant{y}\geqslant\mathrm{0} \\ $$$$\mathrm{does} \\ $$$${a}_{{n}} \geqslant{a}_{{m}} ,\forall{n}\geqslant{m}\geqslant\mathrm{0}? \\ $$ Answered…

Find-f-x-such-that-f-2x-f-x-

Question Number 132932 by bobhans last updated on 17/Feb/21 $${Find}\:{f}\left({x}\right)\:{such}\:{that}\:{f}\left(\mathrm{2}{x}\right)={f}\left({x}\right) \\ $$ Answered by Olaf last updated on 17/Feb/21 $${f}\left(\mathrm{2}{x}\right)\:=\:{f}\left({x}\right) \\ $$$$\Rightarrow\:{f}\left({x}\right)\:=\:{f}\left(\frac{{x}}{\mathrm{2}}\right)\:=\:{f}\left(\frac{{x}}{\mathrm{4}}\right)\:=\:…{f}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)… \\ $$$${f}\left({x}\right)\:=\:\underset{{n}\rightarrow\infty}…

Question-132935

Question Number 132935 by liberty last updated on 17/Feb/21 Answered by EDWIN88 last updated on 17/Feb/21 $$\:\mathrm{let}\:\mathrm{x}=\mathrm{2}\:\Rightarrow\:\mathrm{f}\left(\mathrm{2}\right).\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{f}\left(\mathrm{2}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{65}.\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{65}+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{65}}{\mathrm{64}}\:\Rightarrow\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$\Rightarrow\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{x}^{\mathrm{6}} \\…