y = ln (x e^-x)
y' = (1/(x e^-x) ) * [ x (-e^-x) + e^-x ]
= -1 + 1/x
that derivative is zero when x = 1
then xe^-x = .36788
what is the second derivative?
y'' = 0 - 1/x^2
humm, when x = 1
curvature is - so that is a maximum and the only one I see
I have three questions I'm having a terrible time with:
1)Find, if possible, the absolute maximum value and where it occurs for f(x)=ln(xe^-x) on (0,infinity).
2)Find the value(s) of "c" guaranteed by the Mean Value Theorem for the function f(x)=ln(x^2) on the interval [1,e].
3)Find the derivative of the following function: h(x)=ln(((3x+1)(2)^1/2))((5x^2-4x)^-1)
Thank you!
1 answer