Question

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`

Answer 1

(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]}