For p = 15e^-x, 0 < x < 7, find the local extrema
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
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
To find the local extrema of a function, we need to find the critical points. In order to find the critical points, we take the derivative of the function and set it equal to zero.
First, let's find the derivative of the function p = 15e^(-x). The derivative with respect to x can be found using the chain rule:
dp/dx = d/dx (15e^(-x))
= -15e^(-x)
Next, we set the derivative equal to zero and solve for x:
-15e^(-x) = 0
Since e^(-x) is never equal to zero, the only solution is when -15 equals zero. However, this is not possible, so there are no critical points for this function.
Since there are no critical points, this means there are no local extrema for the function p = 15e^(-x) in the interval 0 < x < 7.