Question-202536

Question Number 202536 by cortano12 last updated on 29/Dec/23

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Answered by taguim001 last updated on 29/Dec/23

f(xy)f(x)−f(xy)f(y)=xf(x)f(y)−yf(x)f(y) {_(f(xy)f(y)=yf(x)f(y)) ^(f(xy)f(x)=xf(x)f(y)) {_(f(xy)=yf(x)) ^(f(xy)=xf(y)) ⇔xf(y)=yf(x) ⇔((f(x))/x)=((f(y))/y) ⇔((f(x))/x)=k (k εR) ⇔f(x)=kx (k εR) f:R→R x ∣→ kx (kεR)

$$ \\ $$$${f}\left({xy}\right){f}\left({x}\right)−{f}\left({xy}\right){f}\left({y}\right)={xf}\left({x}\right){f}\left({y}\right)−{yf}\left({x}\right){f}\left({y}\right) \\ $$$$\left\{_{{f}\left({xy}\right){f}\left({y}\right)={yf}\left({x}\right){f}\left({y}\right)} ^{{f}\left({xy}\right){f}\left({x}\right)={xf}\left({x}\right){f}\left({y}\right)} \right. \\ $$$$\left\{_{{f}\left({xy}\right)={yf}\left({x}\right)} ^{{f}\left({xy}\right)={xf}\left({y}\right)} \right. \\ $$$$\Leftrightarrow{xf}\left({y}\right)={yf}\left({x}\right) \\ $$$$\Leftrightarrow\frac{{f}\left({x}\right)}{{x}}=\frac{{f}\left({y}\right)}{{y}} \\ $$$$\Leftrightarrow\frac{{f}\left({x}\right)}{{x}}={k}\:\:\:\:\left({k}\:\epsilon\mathbb{R}\right) \\ $$$$\Leftrightarrow{f}\left({x}\right)={kx}\:\left({k}\:\epsilon\mathbb{R}\right) \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${x}\:\shortmid\rightarrow\:{kx}\:\:\:\left({k}\epsilon\mathbb{R}\right) \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 30/Dec/23

f(xy)( f(x)−f(y) )=(x−y)f(x)f(y) Searching in linear polynomials Let f(x)=ax+b (axy+b)( ax+b−ay−b )=(x−y)(ax+b)(ay+b) a(axy+b)( x−y )=(x−y)(ax+b)(ay+b) a^2 xy+ab=a^2 xy+abx+aby+b^2 ab=abx+aby+b^2 ab=b^2 ∧ ab=0 b^2 =0⇒b=0 f(x)=ax

$${f}\left({xy}\right)\left(\:{f}\left({x}\right)−{f}\left({y}\right)\:\right)=\left({x}−{y}\right){f}\left({x}\right){f}\left({y}\right) \\ $$$${Searching}\:{in}\:{linear}\:{polynomials} \\ $$$${Let}\:{f}\left({x}\right)={ax}+{b} \\ $$$$\left({axy}+{b}\right)\left(\:{ax}+{b}−{ay}−{b}\:\right)=\left({x}−{y}\right)\left({ax}+{b}\right)\left({ay}+{b}\right) \\ $$$${a}\left({axy}+{b}\right)\left(\:{x}−{y}\:\right)=\left({x}−{y}\right)\left({ax}+{b}\right)\left({ay}+{b}\right) \\ $$$${a}^{\mathrm{2}} {xy}+{ab}={a}^{\mathrm{2}} {xy}+{abx}+{aby}+{b}^{\mathrm{2}} \\ $$$${ab}={abx}+{aby}+{b}^{\mathrm{2}} \\ $$$${ab}={b}^{\mathrm{2}} \:\wedge\:{ab}=\mathrm{0} \\ $$$${b}^{\mathrm{2}} =\mathrm{0}\Rightarrow{b}=\mathrm{0} \\ $$$${f}\left({x}\right)={ax} \\ $$

Answered by witcher3 last updated on 30/Dec/23

f(y)(f(1)−f(y))=(1−y)f(1)f(y) ⇒f(y)(f(1)−f(y)−f(1)(1−y))=0 f(y)=0 or f(y)=f(1)y f(y)=ay f(y)=0 ∈{R→^f R∣∀x∈R ∃! a∈R ∣f(x)=ax} f(y)=ay ;a∈R

$$\mathrm{f}\left(\mathrm{y}\right)\left(\mathrm{f}\left(\mathrm{1}\right)−\mathrm{f}\left(\mathrm{y}\right)\right)=\left(\mathrm{1}−\mathrm{y}\right)\mathrm{f}\left(\mathrm{1}\right)\mathrm{f}\left(\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{y}\right)\left(\mathrm{f}\left(\mathrm{1}\right)−\mathrm{f}\left(\mathrm{y}\right)−\mathrm{f}\left(\mathrm{1}\right)\left(\mathrm{1}−\mathrm{y}\right)\right)=\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{0} \\ $$$$\mathrm{or}\:\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{1}\right)\mathrm{y} \\ $$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{ay} \\ $$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{0}\:\in\left\{\mathbb{R}\overset{\mathrm{f}} {\rightarrow}\mathbb{R}\mid\forall\mathrm{x}\in\mathbb{R}\:\exists!\:\mathrm{a}\in\mathbb{R}\:\mid\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}\right\} \\ $$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{ay}\:;\mathrm{a}\in\mathbb{R} \\ $$

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