document.write( "Question 1158620: Locate the critical points of the following function. Then use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither.
\n" ); document.write( " f(x) = −e^x (x + 6)
\n" ); document.write( "

Algebra.Com's Answer #781692 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "I will presume that you have seen a proof that the product of two differentiable functions is, itself, differentiable, or if not, that you accept this fact on faith. Further, I assume that you accept the fact that and are both differentiable on all reals.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Given the above, it is clear that there is no real number such that does not exist. Hence, the only critical points are where \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "And by inspection, we see that only , hence is the only critical point of the function.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "I leave it as an exercise for you to show that the second derivative is , determine the sign of the value of , and use that result to select local maxima or minima based on:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " Local maxima\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\ 0\ \Right\ \"> Local minima\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "John
\n" ); document.write( "
\n" ); document.write( "My calculator said it, I believe it, that settles it
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" );