Question

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

Answer 1

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.

Hopefully this helps!

Answer 2

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.