Question

In Exercises \(1-6,\) use the First Derivative Test to determine the local extreme values of the function, and identify any absolute extrema. Support your answers graphically. $$y=x e^{1 / x}$$

Short answer

The local maximum is at the point \((-1, - e^{-1})\) and the local minimum is at the point \((1, e^{-1})\). The function does not have any points of absolute extrema.

Step-by-step solution

Compute the First Derivative

Differentiate the given function \(y = xe^{1/x}\) using the product rule for derivatives which states that the derivative of the 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 yields \[ y' = e^{1/x} - xe^{1/x}/x^2 \]

Find Critical Points

Setting the derivative equal to zero allows us to find the critical points that are candidates for local extreme values. Thus, \[ e^{1/x} - xe^{1/x}/x^2 = 0 \] Solving this equation gives \( x = 1 \) and \( x = -1 \).

Apply the First Derivative Test

Plugging in values that are less than -1, between -1 and 1, and greater than 1 into the derivative equation tests the sign. A change from positive to negative indicates a local maximum, while a change from negative to positive indicates a local minimum. Here, the function decreases in \((-∞, -1)\) and \((-1, 1)\) and increases in \((1, +∞)\). Hence, \(x=-1\) is a local maximum and \(x=1\) is a local minimum.

Find the Local Extreme Values

Substitute \( x = -1 \) and \( x = 1 \) into the function to find the local extreme values. The local maximum is \((-1, -e^{-1})\) and the local minimum is \((1, e^{-1})\)

Identify the Absolute Extrema

Because the function decreases without bound as \( x \) approaches negative infinity and increases without bound as \( x \) approaches positive infinity, it does not have any absolute extrema.