f(x) = e^-1/(x + 9)?

Question

find the inflection point, concavity, i dont know how to do this, anyything would help, thanks

Answer 1

F´(x) = e^-1/(x-9) * 1/(x-9)^2
F´´(x) =e^-1/(x-9)* 1/(x-9)^4 +e^-1/(x-9) *(-2)/(x-9)^3=
e^-1/(x-9)*1/(x-9)^3*[1/(x-9)-2]
F´´(x)=0
1-2x+18=0 2x= 19 and x=9.5
sign of F´´ depends on -2x+19

from -infinity to 9.5 positive concavity
from 9.5 to+ infinity negative concavity