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 a local minimum at \(x=-1\).

Step-by-step solution

Find the first derivative

The first derivative of \(y=xe^{x}\) can be found using the product rule, which states that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. This means that \(y^{'}=(1)e^{x} + x(e^{x})= e^{x}+xe^{x}\). Then the critical points of this function are found by setting it equal to zero, which gives \(0=e^{x}+xe^{x}\). Simplifying this equation yields \(x = -1\), which is the only critical point.

Find the second derivative

The second derivative of \(y=xe^{x}\) can also be found using the product rule, which gives \(y^{''}= (1)e^{x} + e^{x} + e^{x} + xe^{x} = 2e^{x}+xe^{x}\). After finding the second derivative, evaluate the function at the critical point \(x = -1\), which gives \(y^{''}(-1) = 2e^{-1} - e^{-1} = e^{-1} > 0\). This tells us that the function is concave up at that critical point.

Apply the Second Derivative Test and interpret the results

The Second Derivative Test states that if a function's second derivative at a critical point is positive, then the function has a local minimum at that point. Therefore, the function \(y=xe^{x}\) has a local minimum at \(x = -1\).