If point of inflection of #y=e^(-x^2)# is at #y=1/e^a#, #a=..........#?
(a) #-1/2#
(b) #-1#
(c) #1#
(d) #1/2#
(a)
(b)
(c)
(d)
1 Answer
(d)
Explanation:
The point of inflection of
So first we need to find the second derivative.
#y=e^(-x^2)#
To differentiate this, first use the chain rule.
#dy/dx=e^(-x^2)(d/dx(-x^2))#
#color(white)(dy/dx)=-2xe^(-x^2)#
Now the second derivative can be found through the product rule:
#(d^2y)/(dx^2)=-2(d/dxx)e^(-x^2)-2x(d/dxe^(-x^2))#
#color(white)((d^2y)/(dx^2))=-2e^(-x^2)-2xe^(-x^2)(d/dx(-x^2))#
#color(white)((d^2y)/(dx^2))=-2e^(-x^2)-2xe^(-x^2)(-2x)#
#color(white)((d^2y)/(dx^2))=4x^2e^(-x^2)-2e^(-x^2)#
#color(white)((d^2y)/(dx^2))=2e^(-x^2)(2x^2-1)#
So, we want to find when this changes sign. Note that
Setting
This means there are points of inflection at both
The
#y(1/sqrt2)=e^(-(1/sqrt2)^2)=e^(-1/2)=1/e^(1/2)#
#y(-1/sqrt2)=e^(-(-1/sqrt2)^2)=e^(-1/2)=1/e^(1/2)#
Thus
graph{e^(-x^2) [-2.5, 2.5, -0.87, 1.63]}