document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167057>Question 1167057</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the inverse function of f(x)=x^2+x, x ≥ -1/2? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #791637 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=791637\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> actually, x needs to be greater than or equal to -1/4, not -1/2, as you will see later.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you let y = f(x), then your function of f(x) = x^2 + x becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y = x^2 + x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since x^2 + x = (x + 1/2)^2 - 1/4, then your function becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y = (x + 1/2)^2 - 1/4\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "add 1/4 to both sides of this function to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y + 1/4 = (x + 1/2)^2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "take the square root of both sides of this function to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "sqrt(y + 1/4) = x + 1/2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "subtract 1/2 from both sides of the function to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "sqrt(y + 1/4) - 1/2 = x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "replace x with y and y with x to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "sqrt(x + 1/4) - 1/2 = y\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "switch sides to get:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y = sqrt(x + 1/4) - 1/2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's your inverse function.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if it is truly the inverse function, then it must be a reflection about the line y = x and (x,y) in the normal function must be the same distance from the line y = x as (y,x) is in the inverse function.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "i graphed both the normal function and the inverse function and the line y = x to show you that the inverse function is a reflection about the line y = x.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "i also graphed the line y = -x + 6 to show you that the point (x,y) in the normal function is the same distance from the line y = x as the point (y,x) in the inverse function.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "here's what the graph looks like.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2020/101106.jpg\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the distance between the point (1.646,4.354) on the normal graph and the point (3,3) on the line y = x is the same distance between the point (4.354,1.646) on the inverse graph and the same point of (3,3) on the line y = x.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "to confirm, use the distance formula of d = sqrt((x2-x1)^2 + (y2-y1)^2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "when (x1,y1) from the normal function = (1.646,4.354) and (x2,y2) from the line (y = x) = (3,3), then:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "d = sqrt((3 - 1.646)^2 + (3 - 4.354)^2) = 1.914845163\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "when (x1,y1) from the inverse function = (4.354,1.646)) and (x2,y2) from the line (y = x) = (3,3), then:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "d = sqrt(3 - 4.354)^2 + (3 - 1.646)^2) = 1.914845163\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the distances are the same, confirming that the inverse function for y = x^2 + x is the function y = sqrt(x + 1/4) - 1/2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );