# If # y=e^(−x^2)#, what are the points of inflection, concavity and critical points?

Given:

we can calculate the first and second order derivatives:

we can therefore determine that:

(1) By solving the equation:

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To find the points of inflection, concavity, and critical points of the function ( y = e^{-x^2} ), we need to take its second derivative and analyze it.

First, find the first derivative of ( y ) with respect to ( x ): [ \frac{dy}{dx} = -2xe^{-x^2} ]

Then, find the second derivative: [ \frac{d^2y}{dx^2} = -2e^{-x^2} + 4x^2e^{-x^2} ]

For points of inflection, set the second derivative equal to zero and solve for ( x ): [ -2e^{-x^2} + 4x^2e^{-x^2} = 0 ] [ e^{-x^2}(4x^2 - 2) = 0 ] [ 4x^2 - 2 = 0 ] [ x^2 = \frac{1}{2} ] [ x = \pm\frac{1}{\sqrt{2}} ]

Now, we need to determine the concavity of the function. For concave up, the second derivative is positive; for concave down, it's negative.

Evaluate the second derivative at ( x = \pm\frac{1}{\sqrt{2}} ): [ \frac{d^2y}{dx^2}\left(\frac{1}{\sqrt{2}}\right) = -2e^{-\frac{1}{2}} + 4\left(\frac{1}{\sqrt{2}}\right)^2e^{-\frac{1}{2}} = -2e^{-\frac{1}{2}} + 2e^{-\frac{1}{2}} = 0 ] [ \frac{d^2y}{dx^2}\left(-\frac{1}{\sqrt{2}}\right) = -2e^{-\frac{1}{2}} + 4\left(-\frac{1}{\sqrt{2}}\right)^2e^{-\frac{1}{2}} = -2e^{-\frac{1}{2}} + 2e^{-\frac{1}{2}} = 0 ]

Therefore, the function ( y = e^{-x^2} ) is concave up and has points of inflection at ( x = \pm\frac{1}{\sqrt{2}} ). To find the critical points, set the first derivative equal to zero and solve for ( x ):

[ -2xe^{-x^2} = 0 ] [ x = 0 ]

Thus, the critical point is at ( x = 0 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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