document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=287526>Question 287526</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find a function f such that the graph of f has a horizontal tangent at (2, 0) and f ''(x) = 4x.\r<BR>\n" );
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document.write( "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 #208328 by </B><A HREF=/tutors/aboutme.mpl?userid=nabla><B>nabla(475)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nabla><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=208328\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Integrate (find the antiderivative) twice:\r<BR>\n" );
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document.write( "f'(x)=2x^2+C<BR>\n" );
document.write( "f(x)=(2/3)x^3+Cx+D\r<BR>\n" );
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document.write( "Now, f has a horizontal tangent at (2,0) IE the function attains a local extrema and f'(2)=0\r<BR>\n" );
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document.write( "hence\r<BR>\n" );
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document.write( "f'(2)=2*2^2+C=0<BR>\n" );
document.write( "implies C=-8.\r<BR>\n" );
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document.write( "Now we have f(x)=(2/3)x^3-8x+D.\r<BR>\n" );
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document.write( "But we know f(2)=0, so, (2/3)2^3-8(2)+D=0<BR>\n" );
document.write( "16/3-48/3=-D<BR>\n" );
document.write( "-32/3=-D\r<BR>\n" );
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document.write( "implies 32/3=D. \r<BR>\n" );
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document.write( "This gives our final function as\r<BR>\n" );
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document.write( "f(x)=(2/3)x^3-8x+32/3. </SPAN>\n" );
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