Question

In Exercises \(33-38\) , use the Second Derivative Test to find the local extrema for the function. $$y=x e^{-x}$$

Short answer

The function \(y = xe^{-x}\) has one local maximum at the point \(x = 1\).

Step-by-step solution

Differentiation

The given function is \(y = xe^{-x}\). To find the first derivative with respect to \(x\), apply the product rule which states: \((fg)' = f'g + fg'\). Applying this to \(y = xe^{-x}\) we get: \[y' = 1*e^{-x} + x*(-e^{-x}) = e^{-x} - xe^{-x}\].

Second Differentiation

The next task is to find the second derivative of the function. This can be done by taking the derivative of \(y'\) from the first step: \[y'' = (e^{-x} - xe^{-x})' = -e^{-x} - e^{-x} + xe^{-x} = -2e^{-x} + xe^{-x}.\]

Apply Second Derivative Test

The Second Derivative Test determines whether the critical points are local minima, local maxima, or saddle points. Setting \(y'=0\):\[e^{-x} - xe^{-x} = 0\]This could be written as \(e^{-x}(1 - x) = 0\) and will be true when \(x = 1\), so the critical point is \(x = 1\). Now, substitute \(x = 1\) into \(y''\), we find \(y''(1) = -2e^{-1} + e^{-1} = -e^{-1} < 0\]. A negative second derivative at a critical point indicates a local maximum.

Conclusion on the Local Extrema

Since the Second Derivative Test confirms that there is a local maximum at \(x = 1\), and there are no other critical points, the only local extremum is a maximum at \(x = 1\).