At what interval is e^(-5x^2) concave up?

I know the second derivative is
100x^2*e^-5x^2-10*e^-5x^2

but I just can not figure this one out.
Thank you for your help!

first

http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29

first

http://www.wolframalpha.com/input/?i=plot++e^%28-5x^2%29

Then this for second derivative and graph

https://www.wolframalpha.com/input/?i=second+derivative+of+e^%28-5x^2%29

Note that you did the second derivative correctly. It is easier to write it their way y" = 10e^(-5x^2)(10x^2-1)

It is zero at x = +/- .316

so it is positive for x>.316

and for x <.316

To determine the intervals on which the function e^(-5x^2) is concave up, we need to examine the sign of the second derivative.

The second derivative you provided is correct: 100x^2e^(-5x^2) - 10e^(-5x^2).

To find the intervals where the second derivative is positive (indicating concave up), we need to solve the inequality:

100x^2e^(-5x^2) - 10e^(-5x^2) > 0

To simplify the equation, we can factor out e^(-5x^2) from both terms:

e^(-5x^2) * (100x^2 - 10) > 0

Now we have two factors: e^(-5x^2) and (100x^2 - 10). To determine when the inequality is true, we analyze the sign of each factor separately.

First, let's consider e^(-5x^2). Since it is an exponential function, it is always positive for any real value of x.

Next, let's analyze (100x^2 - 10). To find the values of x for which this factor is positive, we solve the inequality:

100x^2 - 10 > 0

Adding 10 to both sides:

100x^2 > 10

Dividing both sides by 100:

x^2 > 0.1

Taking the square root of both sides (remembering to consider both the positive and negative roots):

x > √0.1 or x < -√0.1

Now, we can combine the information from both factors:

e^(-5x^2) > 0, for all real values of x
AND
(100x^2 - 10) > 0 for x > √0.1 or x < -√0.1

Combining these conditions, we find that the function e^(-5x^2) is concave up on the intervals (-∞, -√0.1) and (√0.1, +∞).

Therefore, the intervals on which e^(-5x^2) is concave up are (-∞, -√0.1) and (√0.1, +∞).