How do you find the point of inflexion of y=xe^(-x)+3?
Then the inflectional tangent?

Question

#y=xe^-x+3#

Christopher Agresta · Accepted Answer

The inflection point would be #2, 2e^(-2) +3#
The tangent line would then be
#y-(2e^(-2) +3)=-e^(-2) (x-2)# y'=#e^(-x) -xe^(-x)# y"=#-e^(-x) -( e^(-x) -xe^(-x) )# = #xe^(-x) -2e^(-x)# Make y''=0 for the inflection point, which would result in x=2. The inflection point would be #2, 2e^(-2) +3# The slope at this point would be #e^(-2) -2e^(-2)= -e^(-2)# The tangent line would then be #y-(2e^(-2) +3)=-e^(-2) (x-2)#

Charles Arocha · Answer

undefined To find the point of inflection of $ y = xe^{-x} + 3 $, you first need to find the second derivative of the function. Once you have the second derivative, set it equal to zero and solve for $ x $. The value(s) of $ x $ obtained will give you the x-coordinate(s) of the point(s) of inflection. To find the corresponding y-coordinate(s), substitute the $ x $-value(s) into the original function.

The inflectional tangent at the point of inflection can be found by evaluating the first derivative of the function at the x-coordinate(s) of the point(s) of inflection. The equation of the tangent line can be written using the point-slope form, where the slope is given by the value of the first derivative at the point of inflection and the point is the point of inflection itself.

How do you find the point of inflexion of y=xe^(-x)+3? Then the inflectional tangent?

#y=xe^-x+3#