Question

Find the location and value of the absolute extremum for f(x) = -xe^-x/3

Answer 1

f(x)' = -e^(-x/3) + (1/3)xe^(-x/3)
you can factor out the e^(-x/3) giving
e^(-x/3)(-1+(1/3)x)=0
since e is never equal to zero just find when the 2nd function of x = 0
(-1+1/3 x) =0 @ x=3
take the second derivative and get

f(x)'' = 1/3e^-x/3 + 1/3e^-x/3 + 1/3xe^-x/3
again factor out the e and plug in the x=3
e^-1(1/3 +1/3+3/9) = e^-1 which is always positive so the function is concave up.