How do you find the points of inflection of the curve #y=e^(x^2)#?

2 Answers
Mar 19, 2018

Answer:

None, it's concave up for #(-oo, oo)#.

Explanation:

Compute the first derivative.

#y' = 2xe^(x^2)#

Second derivative is given by the product rule.

#y'' = 2(e^(x^2)) + 2x(2x)e^(x^2)#

#y'' = 2e^(x^2) + 4x^2e^(x^2)#

We need to set this to #0# and solve to determine inflection points.

#0 = 2e^(x^2) + 4x^2e^(x^2)#

#0 = 2e^(x^2)(1 + 2x^2)#

We see this has no solution because #2e^(x^2) != 0# for all values of #x#. Furthermore, the second equation states that #1 + 2x^2 = 0 -> 2x^2 = -1 -> x = sqrt(-1/2)#

This doesn't have a real value so no real solution to this equation. This simply means the function #y =e^(x^2)# will have no inflection points (#y''# is positive on all it's domain, therefore concave up on #(-oo, oo)#). We can even confirm graphically.

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Hopefully this helps!

Mar 19, 2018

Answer:

In principle, by differentiating twice, setting the result to zero, and checking whether the result is a genuine point of inflexion. However, this function has no points of inflexion.

Explanation:

Differentiating twice gives #f''(x)=2x*2xe^(x^2)+2*e^(x^2)#
#=2e^(x^2)(2x^2+1)#
This can never be zero because all three terms of the product are always strictly positive

Did you mean #e^(-x^2)#? This has points of inflection when #2x^2-1=0#, that is, #x=±1/sqrt2 #

You should then test that the second derivative changes sign at these points, which it clearly does as #2x^2-1# is a parabola passing through through #(±1/sqrt2,0)#, and the exponential term is always positive.