document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1107577>Question 1107577</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider the following function.\r<BR>\n" );
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document.write( "f&#8201;(x) &#8201;=&#8201; (1/sqrt{4&#960;})*e^(&#8722;x^2/18)<BR>\n" );
document.write( "(note that 4&#960; is under the square root in the denominator) \r<BR>\n" );
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document.write( "Find the largest interval(s) on which &#8201;f&#8201; is concave up. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #722589 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=722589\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Do some analysis first....<br><BR>\n" );
document.write( "(1) The function value is always positive<BR>\n" );
document.write( "(2) The function has its maximum value when x=0<BR>\n" );
document.write( "(3) The function is an even function, symmetric with respect to the y-axis.<BR>\n" );
document.write( "(4) The limit of the function is 0 when x goes to positive or negative infinity<br><BR>\n" );
document.write( "From those observations, we know the graph will be concave down between -a and a for some positive value a and concave up everywhere else.  So there will be two largest intervals where the function is concave up -- from negative infinity to a, and from a to infinity.<br><BR>\n" );
document.write( "We now only need to find the value of a; for simplicity we will work with positive values of x.<br><BR>\n" );
document.write( "The graph changes from concave down to concave up when the second derivative is zero -- that is, when the slope changes from decreasing to increasing.<br><BR>\n" );
document.write( "f(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2Fsqrt%284%2Api%29%29e%5E%28%28-1%2F18%29%2Ax%5E2%29 ALIGN=MIDDLE  ALT=\"%281%2Fsqrt%284%2Api%29%29e%5E%28%28-1%2F18%29%2Ax%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "f'(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2Fsqrt%284%2Api%29%29%28e%5E%28%28-1%2F18%29%2Ax%5E2%29%29%28%28-1%2F9%29x%29+=+%28-x%2F%289%2Asqrt%284%2Api%29%29%29%28e%5E%28%28-1%2F18%29%2Ax%5E2%29%29%28%281%2F9%29x%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >  [chain rule, and derivative of e^f(x)]<br><BR>\n" );
document.write( "f''(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%2F%289%2Asqrt%284%2Api%29%29%29%28e%5E%28%28-1%2F18%29%2Ax%5E2%29%29%2B%28-x%2F%289%2Asqrt%284%2Api%29%29%29%28e%5E%28%28-1%2F18%29%2Ax%5E2%29%29%28%28-1%2F9%29x%29+=+%28e%5E%28%28-1%2F18%29%2Ax%5E2%29%2F%289%2Asqrt%284%2Api%29%29%29%2A%28-1%2Bx%5E2%2F9%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >[product rule, chain rule, and derivative of e^f(x)]<br><BR>\n" );
document.write( "The second derivative is zero when<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2Bx%5E2%2F9+=+0 ALIGN=MIDDLE  ALT=\"-1%2Bx%5E2%2F9+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2F9+=+1 ALIGN=MIDDLE  ALT=\"x%5E2%2F9+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+=+9 ALIGN=MIDDLE  ALT=\"x%5E2+=+9\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+3 ALIGN=MIDDLE  ALT=\"x+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The value of a we are looking for is 3; the two largest intervals on which the graph is concave up are from negative infinity to -3 and from 3 to infinity. </SPAN>\n" );
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