For p = 15e^-x, 0 < x < 7, find the local extrema
4 answers
Sorry, find the local extrema for the revenue function: R = xp
R = 15x e^-x
R' = 15(1-x) e^-x
R'=0 when x=1
min or max?
R" = 15(x-2) e^-x
R"(1) < 0, so R(1)=15 is a maximum
R' = 15(1-x) e^-x
R'=0 when x=1
min or max?
R" = 15(x-2) e^-x
R"(1) < 0, so R(1)=15 is a maximum
When I try to find the first derivative, get R = 15e^-x - 15xe^-x..
yeah, that's what I got, too, if you factor out the 15 e^-x
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