calculus minimum and concavity f(x) = xe^(5/x)?

Question

f(x) = xe^(5/x)

Find the minimum value

Find intervals of concavity. List the intervals without specifying Concave Up vs. Concave Down.

Answer 1

f´(x) = e^5/x -5/x e^5/x = e^5/x(1-5/x)= 0
x/5=1 so x=5 is a critical point and the sign is

++++++ 0(discont)----------- 5 +++++++ so at x=5 you have a local minimum
f(5)=5e This is NOT an absolute minimum as for x=>-infinity
f(x) has limit - infinity

To find concavity call 5/x =z
y´(z)= e^z(1-z) and
dy´/dx =dy´/dz *dz/dx where dz/dx= -5/x^2 always negative

dy´/dz= e^z(1-z)-e^z =-ze^z y´´ =25/x^3e^5/x

sign y´´ ------------0 (disc.)++++++++++