Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 176391 by cortano1 last updated on 18/Sep/22

Commented by mr W last updated on 18/Sep/22

f(x)=x^3

$${f}\left({x}\right)={x}^{\mathrm{3}} \\ $$

Answered by TheHoneyCat last updated on 18/Sep/22

Here′s the proof: yf(2x)−xf(2y)=8xy(x^2 −y^2 ) On (R^∗ )^2 ((f(2x))/x)−((f(2y))/y)=8(x^2 −y^2 )=2((2x)^2 −(2y)^2 ) thus denoting: { ((a=2x)),((b=2y)),((g(x)=((f(x))/x))) :} g(a)−g(b)=a^2 −b^2 fixing b to any value, we get that ∃c∈R/ g: { (R^∗ ,→,R),(x, ,(x^2 +c)) :} now comming back to g(a)−g(b)=a^2 −b^2 fixing b=−a g(a)−g(−a)=0 thus ∀x∈R^∗ g(x)=g(−x) hence c=0 so : g: { (R^∗ ,→,R),(x, ,x^2 ) :} thus f_(∣R^∗ ) : { (R^∗ ,→,R),(x, ,x^3 ) :} the only value left to find is f(0) choosing x=0 and y=1 (or anything such that y≠0) we get that: 1×f(0)−0×f(2)=8×0×1×(0^2 −1^2 ) ie f(0)=0 Thefore, indeed, f : { (R,→,R),(x, ,x^3 ) :} □

$$\mathrm{Here}'\mathrm{s}\:\mathrm{the}\:\mathrm{proof}: \\ $$$$ \\ $$$${yf}\left(\mathrm{2}{x}\right)−{xf}\left(\mathrm{2}{y}\right)=\mathrm{8}{xy}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{On}\:\left(\mathbb{R}^{\ast} \right)^{\mathrm{2}} \\ $$$$\frac{{f}\left(\mathrm{2}{x}\right)}{{x}}−\frac{{f}\left(\mathrm{2}{y}\right)}{{y}}=\mathrm{8}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{2}\left(\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\left(\mathrm{2}{y}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{thus}\:\mathrm{denoting}:\:\begin{cases}{{a}=\mathrm{2}{x}}\\{{b}=\mathrm{2}{y}}\\{{g}\left({x}\right)=\frac{{f}\left({x}\right)}{{x}}}\end{cases} \\ $$$${g}\left({a}\right)−{g}\left({b}\right)={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\mathrm{fixing}\:{b}\:\mathrm{to}\:\mathrm{any}\:\mathrm{value},\:\mathrm{we}\:\mathrm{get}\:\mathrm{that} \\ $$$$\exists{c}\in\mathbb{R}/\:\:\:{g}:\:\begin{cases}{\mathbb{R}^{\ast} }&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{{x}^{\mathrm{2}} +{c}}\end{cases} \\ $$$$\mathrm{now}\:\mathrm{comming}\:\mathrm{back}\:\mathrm{to}\:{g}\left({a}\right)−{g}\left({b}\right)={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\mathrm{fixing}\:{b}=−{a} \\ $$$${g}\left({a}\right)−{g}\left(−{a}\right)=\mathrm{0} \\ $$$$\mathrm{thus}\:\forall{x}\in\mathbb{R}^{\ast} \:\:\:{g}\left({x}\right)={g}\left(−{x}\right) \\ $$$$\mathrm{hence}\:{c}=\mathrm{0} \\ $$$$\mathrm{so}\:: \\ $$$$\:\:{g}:\:\begin{cases}{\mathbb{R}^{\ast} }&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{{x}^{\mathrm{2}} }\end{cases} \\ $$$$ \\ $$$$\mathrm{thus}\:{f}_{\mid\mathbb{R}^{\ast} } \::\:\begin{cases}{\mathbb{R}^{\ast} }&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{{x}^{\mathrm{3}} }\end{cases} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{only}\:\mathrm{value}\:\mathrm{left}\:\mathrm{to}\:\mathrm{find}\:\mathrm{is}\:{f}\left(\mathrm{0}\right) \\ $$$$\mathrm{choosing}\:{x}=\mathrm{0}\:\mathrm{and}\:{y}=\mathrm{1}\:\left(\mathrm{or}\:\mathrm{anything}\:\mathrm{such}\:\mathrm{that}\:{y}\neq\mathrm{0}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{that}: \\ $$$$\mathrm{1}×{f}\left(\mathrm{0}\right)−\mathrm{0}×{f}\left(\mathrm{2}\right)=\mathrm{8}×\mathrm{0}×\mathrm{1}×\left(\mathrm{0}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right) \\ $$$${ie}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Thefore},\:\mathrm{indeed},\: \\ $$$${f}\::\:\begin{cases}{\mathbb{R}}&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{{x}^{\mathrm{3}} }\end{cases} \\ $$$$\Box \\ $$

Answered by mr W last updated on 19/Sep/22

let y=(x/2) ((xf(2x))/2)−xf(x)=4x^2 ×((3x^2 )/4) f(2x)−2f(x)=6x^3 let f(x)=ax^3 8ax^3 −2ax^3 =6x^3 ⇒a=1 ⇒f(x)=x^3

$${let}\:{y}=\frac{{x}}{\mathrm{2}} \\ $$$$\frac{{xf}\left(\mathrm{2}{x}\right)}{\mathrm{2}}−{xf}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} ×\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${f}\left(\mathrm{2}{x}\right)−\mathrm{2}{f}\left({x}\right)=\mathrm{6}{x}^{\mathrm{3}} \\ $$$${let}\:{f}\left({x}\right)={ax}^{\mathrm{3}} \\ $$$$\mathrm{8}{ax}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{3}} =\mathrm{6}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={x}^{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 20/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com