Question

In Exercises \(51-54,\) find any asymptotes and relative extrema that may exist and use a graphing utility to graph the function. (Hint: Some of the limits required in finding asymptotes have been found in previous exercises.) \(y=2 x e^{-x}\)

Short answer

The given function has a horizontal asymptote at \(y=0\) and a relative minimum at \(x=2\).

Step-by-step solution

Finding the Asymptotes

To find the asymptotes, calculate the limit of the function as \(x\) approaches to positive and negative infinity.\n\nAs \(x\) approaches to positive infinity (\(x \rightarrow \infty\)), the more negative \(e^-x\) will be, causing the function to approach zero. Hence, the horizontal asymptote is \(y=0\).\n\nThe function has no vertical asymptotes because \(e^-x\) is never equal to zero.

Finding relative extrema

The relative extrema are obtained by taking the derivative of the function and set it to zero. Use product rule: \(d(uv) = v du + u dv\), where \(u=2x\) and \(v=e^-x\). So, \(dy/dx = e^-x(2-x)\). Then solving the equation \(2-x=0\) will give \(x=2\). This \(x\) value will be tested in the second derivative to determine whether it is a maximum or minimum.

Determining nature of extrema

The sign of the second derivative at \(x=2\) will reveal whether it is a maximum or minimum. Taking second derivative yields \(d^2y/dx^2 = e^-x(-x)\). Substituting \(x=2\) into the second derivative yields a positive number so, \(x=2\) is a relative minimum for the function.

Graphing the function

Using a graphing tool, scale the graph properly. Draw the horizontal asymptote \(y=0\). Then, plot the relative minimum at \(x=2\). Sketch the graph rising from the left, reaching the relative minimum, and then getting closer to the asymptote.