Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. Determine whether or not \(f\) attains an absolute maximum and absolute minimum value. If it does, determine the absolute maximum and/or minimum value. \(f(x)=\frac{e^{x}}{2 x}\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{e^{x}}{2x}\) has one critical point at \(x=2\) which is a local maximum. The function has an absolute maximum at \(x=2\) and is undefined at \(x=0\). It does not have an absolute minimum.

Step 1: Identify critical points

Differentiate the function \(f(x)\), set the derivative equal to zero and solve for \(x\). The derivative of \(f(x) = \frac{e^{x}}{2 x}\) is \(f'(x)=\frac{x e^{x}-2 e^{x}}{4 x^{2}}\). Equating this to zero gives \(x - 2 = 0\), therefore, \(x=2\) is the only critical point.

Step 2: Classify critical points

To classify the critical point, take the second derivative of \(f(x)\). The second derivative of \(f(x)\) is \(f''(x) = \frac{-x^{2} e^{x}+4 x e^{x}-8 e^{x}}{8 x^{3}}\). Substituting \(x = 2\) into \(f''(x)\) gives \(f''(2) < 0\), so the critical point \(x=2\) is a local maximum.

Step 3: Determine absolute maximum and minimum values

Since the function is defined for \(x > 0\), consider the function values at endpoints and at the critical point. From the calculation, we get \(f(0) = \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{e^{x}}{2 x}\) which essentially is undefined, and \(f(2) = \frac{e^{2}}{4}\). Therefore, since the function is undefined at \(x = 0\) and reaches a local maximum at \(x = 2\), \(f\) has a global maximum at \(x = 2\). As \(x\) approaches \(+\infty\), the function \(\frac{e^{x}}{2 x}\) approaches \(+\infty\), implying that the function has no absolute minimum.

Key Concepts

absolute maximum

When working with functions, it's important to find the absolute maximum to understand where the function reaches its highest value. For the function given in the exercise, namely \(f(x)=\frac{e^{x}}{2 x}\), finding the absolute maximum involves identifying if there's a point where the function reaches its highest height.

• The exercise found that \(x = 2\) is a critical point.
• By evaluating the function at this point, \(f(2) = \frac{e^{2}}{4}\), we identify the highest value the function attains within the specified domain of \(x > 0\).
• Since the function is undefined at \(x = 0\) and diverges to infinity as \(x\) approaches infinity, no other points can exceed the value at \(x = 2\).

Thus, the function's absolute maximum is at \(x = 2\) with the value \(f(2) = \frac{e^2}{4}\). Keep in mind, finding absolute maxima helps in understanding the range and behavior of functions especially when dealing with constraints.

absolute minimum

An absolute minimum is the point in the domain of a function where it achieves its lowest value. In this exercise, we looked to determine if such a point exists for the function \(f(x)=\frac{e^{x}}{2 x}\).

• Unlike the maximum, which was observed at \(x = 2\), the behavior of the function as \(x\) approaches infinity is crucial.
• The function \(\frac{e^{x}}{2x}\) tends to grow without bound as \(x\) increases.
• Near \(x = 0\), the function is undefined, which means there's no starting point from zero, eliminating the possibility of a minimum there.

Therefore, there is no absolute minimum for this function within its given domain. When analyzing functions, understanding both absolute maxima and minima provides insight into the behavioral extremes the function can achieve across its domain.

second derivative test

The second derivative test is a helpful tool for classifying critical points of a function, to determine what type of extremum they might be.

• For the function \(f(x)=\frac{e^{x}}{2x}\), we identified \(x=2\) as a critical point using the first derivative.
• We then calculate the second derivative: \(f''(x) = \frac{-x^{2} e^{x}+4 x e^{x}-8 e^{x}}{8 x^{3}}\).
• Substituting \(x=2\) into \(f''(x)\), we found \(f''(2) < 0\).

This result indicates a local maximum at \(x=2\) because a negative second derivative typically corresponds to a concave down curve at that point. Therefore, the second derivative test confirms that \(x=2\) is indeed where a local maximum occurs. Remember, the second derivative test is an efficient way to classify critical points and understand the nature of extremums in a function.