document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209949>Question 1209949</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The function f : \mathbb{R} \rightarrow \mathbb{R} satisfies<BR>\n" );
document.write( "f(x) f(y) - f(xy) = -2x - 6y + 10<BR>\n" );
document.write( "for all x, y \in \mathbb{R}. Find f(x).  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851179 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851179\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the given equation be<BR>\n" );
document.write( "$$f(x)f(y) - f(xy) = -2x - 6y + 10 \quad (*)$$<BR>\n" );
document.write( "for all $x, y \in \mathbb{R}$.\r<BR>\n" );
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document.write( "Let $x = 0$ in (*):<BR>\n" );
document.write( "$$f(0)f(y) - f(0) = -6y + 10$$<BR>\n" );
document.write( "$$f(y)f(0) = -6y + f(0) + 10$$\r<BR>\n" );
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document.write( "Let $y = 0$ in (*):<BR>\n" );
document.write( "$$f(x)f(0) - f(0) = -2x + 10$$<BR>\n" );
document.write( "$$f(x)f(0) = -2x + f(0) + 10$$\r<BR>\n" );
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document.write( "Since $f(x)f(0) = -2x + f(0) + 10$ and $f(y)f(0) = -6y + f(0) + 10$, we can substitute $x = 0$ into the latter equation, and $y = 0$ into the former equation:\r<BR>\n" );
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document.write( "If $x = 0$ and $y = 0$:<BR>\n" );
document.write( "$$f(0)f(0) - f(0) = 10$$<BR>\n" );
document.write( "$$f(0)^2 - f(0) - 10 = 0$$<BR>\n" );
document.write( "Let $f(0) = a$. Then $a^2 - a - 10 = 0$.\r<BR>\n" );
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document.write( "If $f(0) = a$, then<BR>\n" );
document.write( "$$f(x)a = -2x + a + 10 \implies f(x) = \frac{-2x + a + 10}{a}$$<BR>\n" );
document.write( "$$f(y)a = -6y + a + 10 \implies f(y) = \frac{-6y + a + 10}{a}$$\r<BR>\n" );
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document.write( "Substituting into the original equation (*):<BR>\n" );
document.write( "$$\left(\frac{-2x+a+10}{a}\right)\left(\frac{-6y+a+10}{a}\right) - \left(\frac{-2xy+a+10}{a}\right) = -2x - 6y + 10$$<BR>\n" );
document.write( "$$\frac{(-2x+a+10)(-6y+a+10)}{a^2} - \frac{-2xy+a+10}{a} = -2x - 6y + 10$$<BR>\n" );
document.write( "$$(-2x+a+10)(-6y+a+10) - a(-2xy+a+10) = a^2(-2x - 6y + 10)$$<BR>\n" );
document.write( "$$12xy - 2x(a+10) - 6y(a+10) + (a+10)^2 + 2axy - a(a+10) = a^2(-2x - 6y + 10)$$<BR>\n" );
document.write( "$$12xy - 2ax - 20x - 6ay - 60y + a^2 + 20a + 100 + 2axy - a^2 - 10a = -2a^2x - 6a^2y + 10a^2$$<BR>\n" );
document.write( "$$12xy + 2axy - 2ax - 20x - 6ay - 60y + 10a + 100 = -2a^2x - 6a^2y + 10a^2$$\r<BR>\n" );
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document.write( "Comparing the terms involving $xy$:<BR>\n" );
document.write( "$$12 + 2a = 0$$<BR>\n" );
document.write( "$$2a = -12$$<BR>\n" );
document.write( "$$a = -6$$\r<BR>\n" );
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document.write( "Substituting $a = -6$:<BR>\n" );
document.write( "$$12xy - 12xy - 20x + 36x - 60y + 36y - 60 + 100 = -72x + 216y + 360$$<BR>\n" );
document.write( "$$16x - 24y + 40 = -72x + 216y + 360$$<BR>\n" );
document.write( "$$88x - 240y - 320 = 0$$<BR>\n" );
document.write( "$$11x - 30y - 40 = 0$$\r<BR>\n" );
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document.write( "This is not true for all x and y. So, there must be a mistake.\r<BR>\n" );
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document.write( "Let's assume $f(x) = Ax + B$. Then<BR>\n" );
document.write( "$$(Ax + B)(Ay + B) - Axy - B = -2x - 6y + 10$$<BR>\n" );
document.write( "$$A^2xy + ABx + ABy + B^2 - Axy - B = -2x - 6y + 10$$<BR>\n" );
document.write( "$$(A^2 - A)xy + ABx + ABy + B^2 - B = -2x - 6y + 10$$\r<BR>\n" );
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document.write( "Comparing the coefficients:<BR>\n" );
document.write( "$$A^2 - A = 0 \implies A(A-1) = 0 \implies A = 0 \text{ or } A = 1$$<BR>\n" );
document.write( "If $A = 0$, then $B^2 - B = -2x - 6y + 10$, which is impossible.<BR>\n" );
document.write( "If $A = 1$, then $Bx + By + B^2 - B = -2x - 6y + 10$.<BR>\n" );
document.write( "Comparing coefficients:<BR>\n" );
document.write( "$$B = -2, B = -6, B^2 - B = 10$$\r<BR>\n" );
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document.write( "This has no solution.\r<BR>\n" );
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document.write( "Let's check $f(x) = 2x+2$<BR>\n" );
document.write( "$(2x+2)(2y+2) - 2xy-2 = 4xy+4x+4y+4 - 2xy -2 = 2xy + 4x + 4y + 2$.\r<BR>\n" );
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document.write( "Let's check $f(x) = Ax + B$.<BR>\n" );
document.write( "If $f(x)=2x+10$:<BR>\n" );
document.write( "$(2x+10)(2y+10) - 2xy -10 = 4xy+20x+20y+100-2xy-10=2xy+20x+20y+90$.\r<BR>\n" );
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document.write( "Let $y = 1$:<BR>\n" );
document.write( "$f(x)f(1) - f(x) = -2x - 6 + 10 = -2x + 4$<BR>\n" );
document.write( "$f(x)(f(1)-1) = -2x + 4$\r<BR>\n" );
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document.write( "Let $x=1$:<BR>\n" );
document.write( "$f(1)f(y) - f(y) = -2 - 6y + 10 = -6y + 8$<BR>\n" );
document.write( "$f(y)(f(1)-1) = -6y + 8$\r<BR>\n" );
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document.write( "Then $f(x) = \frac{-2x+4}{f(1)-1}$ and $f(y) = \frac{-6y+8}{f(1)-1}$.<BR>\n" );
document.write( "Then $\frac{-2x+4}{f(1)-1} = \frac{1}{3} \frac{-6y+8}{f(1)-1}$.<BR>\n" );
document.write( "Then $3(-2x+4) = -6y+8$.<BR>\n" );
document.write( "$-6x+12=-6y+8$, which is not true.\r<BR>\n" );
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document.write( "If $f(x) = 2x+10$, then $f(x)f(y)-f(xy) = (2x+10)(2y+10) - (2xy+10)= 4xy+20x+20y+100 -2xy -10 = 2xy+20x+20y+90$\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{2x+10}$<BR>\n" );
document.write( " </SPAN>\n" );
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