Question

In Exercises \(11-18,\) use analytic methods to find the extreme values of the function on the interval and where they occur. $$k(x)=e^{-x^{2}}, \quad-\infty< x <\infty$$

Short answer

The function \(k(x) = e^{-x^2}\) has a relative minimum at \(x = 0\).

Step-by-step solution

Compute the First Derivative

Differentiate \(k(x) = e^{-x^2}\) with respect to \(x\). Using the chain rule, the derivative is \(k'(x) = -2x * e^{-x^2}\).

Identify Critical Numbers

Set the first derivative equal to zero and solve for \(x\). It gives \(0 = -2x * e^{-x^2}\). Since \(e^{-x^2}\) is never zero, it must be that \(x = 0\). The only critical point is \(x = 0\).

Compute the Second Derivative

Differentiate the derivative \(k'(x) = -2x * e^{-x^2}\) with respect to \(x\) to get the second derivative. It is \(k''(x) = 2*(e^{-x^2} - 2x^2 * e^{-x^2})\).

Apply the Second Derivative Test

Determine the sign of the second derivative for the critical point. It gives \(k''(0) = 2*(e^{0} - 0^2 * e^{0})\), which equals 2. As this is positive, the second derivative test suggests that \(k(x)\) has a relative minimum at \(x = 0\).