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#

1 Answer

(d) #a=1/2#

Explanation:

The point of inflection of #y# is where the second derivative #(d^2y)/(dx^2)# changes sign (from positive to negative or vice versa).

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 #e^(-x^2)>0# for #x inRR#, so the sign of the second derivative is dependent upon the sign of #2x^2-1#.

Setting #2x^2-1=0# and examining the signs around these points reveals that #(d^2y)/(dx^2)<0# on #x in(-1/sqrt2,1/sqrt2)# and #(d^2y)/(dx^2)>0# on #x in(-oo,-1/sqrt2)uu(1/sqrt2,oo)#.

This means there are points of inflection at both #x=-1/sqrt2# and #x=1/sqrt2#.

The #y# coordinates for these points are identical:

#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 #a=1/2# and the correct answer is (d).
graph{e^(-x^2) [-2.5, 2.5, -0.87, 1.63]}