What are the points of inflection, if any, of #f(x)=2e^(-x^2) #?

Question

Emilia Aguilar · Accepted Answer

#f(x) = 2e^(-x^2)# has two inflection points for #x=+-1/sqrt(2)#

A necessary condition for #f(x)# to have an inflection point in #x=barx# is that:

#f''(bar x)=0#

Thus, we determine:

#f'(x) =-4xe^(-x^2)# #f''(x)=-4e^(-x^2)+8x^2e^(-x^2)=e^(-x^2)(8x^2-4)#

As the exponential is always positive, #f''(x)# is null when:

#8x^2-4=0#

That is:

#x=+-1/sqrt(2)#

To be sure that these are inflection points for #f(x)# we must check that #f''(x)# changes sign around its zeroes: this is certainly the case, because #8x^2-4=0# is a second degree polynomial and changes sign on the two sides of single order zeroes.

plot{2e^(-x^2) [-2.54, 2.46, -2.26, 0.24]}

Isaiah Allen · Answer

undefined To find the points of inflection, we need to find the second derivative of the function.

The second derivative of f(x) = 2e^(-x^2) is: f''(x) = 4x^2 e^(-x^2) - 2e^(-x^2).

Next, we set the second derivative equal to zero and solve for x: 4x^2 e^(-x^2) - 2e^(-x^2) = 0.

Factoring out e^(-x^2), we get: e^(-x^2)(4x^2 - 2) = 0.

This equation equals zero when e^(-x^2) = 0 or when 4x^2 - 2 = 0.

However, e^(-x^2) is never zero, so we only need to solve 4x^2 - 2 = 0.

Solving for x gives x = ±√(2/4) = ±√(1/2) = ±(1/√2).

Therefore, the points of inflection are at (±(1/√2), f(1/√2)).