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For what values of x is the graph of y=e−x2 concave down? Provide the values as an interval or list of intervals. To find this, you'll need to determine where the second derivative is negative

User Marsenau
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Final answer:

The graph of y=e^(-x^2) is concave down for values of x in the interval (-∞, -sqrt(0.5)) and (sqrt(0.5), ∞).

Step-by-step explanation:

To determine the values of x for which the graph of y=e^(-x^2) is concave down, we need to find where the second derivative is negative.

Let's start by finding the first and second derivatives of the function.

The first derivative of y=e^(-x^2) is:

dy/dx = -2x * e^(-x^2)

The second derivative is:

d^2y/dx^2 = -2e^(-x^2) + 4x^2 * e^(-x^2)

To find where the second derivative is negative, we need to solve the inequality -2e^(-x^2) + 4x^2 * e^(-x^2) < 0.

Simplifying this inequality gives us: -2 + 4x^2 < 0.

Solving for x^2 gives x^2 < 0.5, which means x < sqrt(0.5) or x > - sqrt(0.5).

Therefore, the graph is concave down for values of x in the interval (-∞, -sqrt(0.5)) and (sqrt(0.5), ∞).

User Ryan Weiss
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