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Question Number 25023 by Tinkutara last updated on 01/Dec/17
Consider the function f(x) which satisfying the functional equation 2f(x) + f(1 − x) = x^2 + 1, ∀ x ∈ R and g(x) = 3f(x) + 1. The range of φ(x) = g(x) + (1/(g(x) + 1)) is
$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{which} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{functional}\:\mathrm{equation} \\ $$$$\mathrm{2}{f}\left({x}\right)\:+\:{f}\left(\mathrm{1}\:−\:{x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{1},\:\forall\:{x}\:\in\:{R} \\ $$$$\mathrm{and}\:{g}\left({x}\right)\:=\:\mathrm{3}{f}\left({x}\right)\:+\:\mathrm{1}.\:\mathrm{The}\:\mathrm{range}\:\mathrm{of} \\ $$$$\phi\left({x}\right)\:=\:{g}\left({x}\right)\:+\:\frac{\mathrm{1}}{{g}\left({x}\right)\:+\:\mathrm{1}}\:\mathrm{is} \\ $$
Answered by prakash jain last updated on 01/Dec/17
2f(x) + f(1 − x) = x^2 + 1 (1) substitue x by 1−x 2f(1−x)+f(x)=(1−x)^2 +1 4f(x) + 2f(1 − x) = 2x^2 + 2 (multiply (1) by 2) subtract 3f(x)=2x^2 −(1−x)^2 +1 3f(x)=2x^2 −1+2x−x^2 +1 =x^2 +2x g(x)=3f(x)+1=(x+1)^2 ∅(x)=(x+1)^2 +(1/((x+1)^2 +1)) I think you can calculate range now.
$$\mathrm{2}{f}\left({x}\right)\:+\:{f}\left(\mathrm{1}\:−\:{x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{substitue}\:{x}\:{by}\:\mathrm{1}−{x} \\ $$$$\mathrm{2}{f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{4}{f}\left({x}\right)\:+\:\mathrm{2}{f}\left(\mathrm{1}\:−\:{x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{2}\:\:\left({multiply}\:\left(\mathrm{1}\right)\:{by}\:\mathrm{2}\right) \\ $$$$\mathrm{subtract} \\ $$$$\mathrm{3}{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{3}{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}−{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${g}\left({x}\right)=\mathrm{3}{f}\left({x}\right)+\mathrm{1}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\emptyset\left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{calculate}\:\mathrm{range}\:\mathrm{now}. \\ $$
Commented by Tinkutara last updated on 02/Dec/17
I already have solved upto this. I have doubt in finding range only.
$$\mathrm{I}\:\mathrm{already}\:\mathrm{have}\:\mathrm{solved}\:\mathrm{upto}\:\mathrm{this}.\:\mathrm{I} \\ $$$$\mathrm{have}\:\mathrm{doubt}\:\mathrm{in}\:\mathrm{finding}\:\mathrm{range}\:\mathrm{only}. \\ $$
Commented by prakash jain last updated on 02/Dec/17
for x∈R Range of φ(x)=[1,∞)
$$\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{Range}\:\mathrm{of}\:\phi\left({x}\right)=\left[\mathrm{1},\infty\right) \\ $$

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