document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=643090>Question 643090</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Could you please help me with this problem? I am given the function f(x)= x^2 + 2x + 6 with a domain where x = all real numbers, and x is &#8804; k. F is a one to one function, and I am to determine the greatest possible value of k. When k has that value, I am then to determine the range of f, and then the inverse function f^-1 and state its domain and range. After that I am to graph f(x) and f^-1(x). I believe I know how to find the value of k; it's just asking for the upper limit of the domain of the function given. I think to determine the range you can complete the square and put the equation into standard parabola form and so determine the vertex from that, using the y point of the vertex and the fact that it is either an up/down parabola (I believe this one is up) to find the range. I'm just having trouble finding the inverse function. When I switch the x's and y's from the original equation and try to solve for y I seem to get stuck. I know how to find the domain and range from there and I am fairly certain I can graph the equation. I just need help finding the inverse function. Please help? :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #404366 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=404366\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "In order to get the range of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE f\"> you first have to establish the maximum value of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\">.  The key to this is the fact that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE f\"> has been declared a 1 to 1 function.  Parabolas, in their full glory, are never 1 to 1 functions because for any value of the independent variable there is a second value of the independent variable on the other side of and equidistant from the vertex that gives the same function value.  In other words a full parabola never passes the horizontal line test.  Enter this mysterious value <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\">.  If you think about it a second, you can choose a value for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\"> at or less than the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\">-coordinate of the vertex and you will guarantee yourself a 1 to 1 function.  Obviously, if you choose <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\ =\ x_v\">, then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\"> will be at its maximum value.\r<BR>\n" );
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document.write( "So where is the vertex?  For any quadratic function in standard form, i.e., <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c\">, the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\">-coordinate of the vertex is located at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x_v\ =\ \frac{-b}{2a}\">, and the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE y\">-coordinate is just the value of the function at that value.\r<BR>\n" );
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document.write( "Let's find your vertex:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ \frac{-2}{2}\ =\ -1\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ f(x_v)\ =\ (-1)^2\ +\ 2(-1)\ +\ 6\ =\ 5\">\r<BR>\n" );
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document.write( "Hence, your vertex is at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE (-1,5)\">\r<BR>\n" );
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document.write( "Now we know that the maximum possible value of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE k\"> is -1, and, since we know the parabola opens upward because of the positive lead coefficient on <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE f\">, the value of the function at the vertex is a minimum, we can say that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \text{ran}(f)\ =\ \left{y\,\in\,\mathbb{R}\,:\,y\ \geq\ 5\right}\">\r<BR>\n" );
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document.write( "On to the inverse.  Let's discover the inverse relation that would derive from the original function.  As is normal, replace <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE f(x)\"> with <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE y\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ 2x\ +\ 6\">\r<BR>\n" );
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document.write( "Now let's go about solving for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\"> in terms of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE y\">.  Put all the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE x\"> terms in the LHS, and every thing else in the RHS.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ =\ y\ -\ 6\">\r<BR>\n" );
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document.write( "Complete the square on the LHS:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ =\ y\ -\ 5\">\r<BR>\n" );
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document.write( "Factor the LHS perfect square:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^2\ =\ y\ -\ 5\">\r<BR>\n" );
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document.write( "Take the root:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ \pm\sqrt{y\ -\ 5}\">\r<BR>\n" );
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document.write( "Finish up\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1\ \pm\sqrt{y\ -\ 5}\">\r<BR>\n" );
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document.write( "And finally, swap the variables:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \ y\ =\ -1\ \pm\sqrt{x\ -\ 5}\">\r<BR>\n" );
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document.write( "Almost there but we aren't quite ready to call this an inverse function yet -- because it is not a function. We have both sides of the \"on its side\" parabola and we only want one of the sides to match our original function.  Since our original function was the left side of the parabola, the inverse has to be the bottom side of the resulting sideways parabola (because of the required symmetry of a function and its inverse about the line <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE y\ =\ x\">), hence:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ f^{-1}(x)\ =\ y\ =\ -1\ -\ \sqrt{x\ -\ 5}\">\r<BR>\n" );
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document.write( "Swapping the coordinates of the vertex on the original function will give you the vertex on the inverse.  Your domain and range of the inverse should be easy to see from there.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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