Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=\sqrt{x} \ln x \text { on }(0, \infty)$$

Short answer

Answer: The absolute minimum value of the function is \(-\frac{2}{e}\) at \(x=e^{-2}\).

Step-by-step solution

Find the critical points

To find the critical points, first find the first derivative of the function. To do that, use the product rule. Let \(u=\sqrt{x}\) and \(v=\ln{x}\), so we can write \(f(x)=uv\). Now find the derivatives of \(u\) and \(v\) with respect to \(x\) and then apply the product rule: $$u'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}, \quad v'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$$ Now, use the product rule: $$f'(x) = u'v + uv' = \frac{1}{2\sqrt{x}}\ln{x} + \frac{\sqrt{x}}{x}$$ To find the critical points, set \(f'(x)\) equal to 0: $$0 = \frac{1}{2\sqrt{x}}\ln{x} + \frac{\sqrt{x}}{x}$$

Apply the First Derivative Test

Now apply the First Derivative Test to determine if the critical points correspond to local maxima or minima. Unfortunately, there isn't an obvious way to solve \(f'(x)=0\) algebraically, so we need to analyze the behavior of the first derivative without finding the explicit critical points. Notice that \(\frac{1}{2\sqrt{x}}\ln{x}\) is positive for values of \(x>e^{-2}\) and negative for \(00\). To study the behavior of \(f'(x)\) within the interval, consider values of \(x\) separated by \(e^{-2}\): - For \(0e^{-2}\), \(f'(x) = \frac{1}{2\sqrt{x}}\ln{x} + \frac{\sqrt{x}}{x} > 0\), which means \(f(x)\) is increasing. There is a change in the sign of the first derivative at \(x=e^{-2}\). Thus, \(x=e^{-2}\) is a critical point and corresponds to a local minimum, as the function changes from decreasing to increasing.

Identify the absolute maximum and minimum values

The function is defined on \((0,\infty)\), so there are no endpoints to consider for absolute maxima or minima, only the critical point found in the previous step. As \(f(x)\) is decreasing for \(0e^{-2}\), it can be inferred that the function has an absolute minimum at \(x=e^{-2}\). Hence, the absolute minimum value is \(f(e^{-2})=\sqrt{e^{-2}}\ln{(e^{-2})}=\frac{1}{\sqrt{e^2}}(-2)=-\frac{2}{e}\). There is no absolute maximum value in the interval \((0,\infty)\), as the function keeps increasing indefinitely.

Key concepts

First Derivative Test

The First Derivative Test is a fundamental tool in calculus used to determine whether critical points of a function are local maxima, minima, or neither. Once you've calculated the first derivative, you set it equal to zero to find critical points, which are potential candidates for these local extremes.

To apply the First Derivative Test, examine the sign changes of the derivative around these critical points. Here's how it works:

• If the first derivative changes from positive to negative at a critical point, the function has a local maximum there.
• If the derivative changes from negative to positive, there is a local minimum.
• If there's no sign change, the critical point is not an extreme point.

In our exercise, for the function \(f(x) = \sqrt{x} \ln x\), the analysis shows a change from negative to positive, indicating a local minimum at \(x = e^{-2}\). This insightful test allows us to conclude about the local behavior of functions without needing graphical plots.

Absolute Extremes

Identifying absolute extremes involves not only checking critical points but also evaluating the behavior over the entire domain of the function. When dealing with absolute extrema, you’re looking for the highest and lowest points a function achieves on a specified interval.

To find the absolute maxima or minima:

• Evaluate the function at critical points found within the domain.
• If the function is defined on a closed interval, also check the endpoint values. However, our function \(f(x) = \sqrt{x} \ln x\) deals with an open interval \((0, \infty)\), so endpoint evaluation isn't applicable.
• Compare these values to determine the highest and lowest values that the function reaches.

For the function \(f(x)\), the absolute minimum occurs at the local minimum \(x = e^{-2}\), where the value is \(-\frac{2}{e}\). There is no absolute maximum since the function increases without bound as \(x\to\infty \).

Product Rule

The Product Rule is a key differentiation rule that allows us to find the derivative of functions that are products of two other functions. Given two functions \(u(x)\) and \(v(x)\), their derivative is given by:

• \(f'(x) = u'(x)v(x) + u(x)v'(x)\)

Useful when dealing with functions like \(f(x) = \sqrt{x} \ln x\), which is essentially a product of \(u(x) = \sqrt{x}\) and \(v(x) = \ln x\). By differentiating \(u\) and \(v\) individually:

• \(u'(x) = \frac{1}{2\sqrt{x}}\)
• \(v'(x) = \frac{1}{x}\)

And then applying the product rule, you obtain the first derivative \(f'(x) = \frac{1}{2\sqrt{x}}\ln{x} + \frac{\sqrt{x}}{x}\). This derivative is fundamental to finding critical points and analyzing function behavior.

Derivative Analysis

Derivative analysis encompasses techniques to understand functions' behaviors through derivatives. It involves finding critical points, evaluating the increasing or decreasing nature of a function, and assessing concavity by using first and second derivatives.

In our exercise, the first derivative \(f'(x) = \frac{1}{2\sqrt{x}}\ln{x} + \frac{\sqrt{x}}{x}\) was derived, enabling us to note:

• The positivity or negativity of the first derivative tells us about the function's increasing or decreasing trends across regions of the domain.
• Identifying the sign changes at key points help in pinpointing local maxima and minima.

This careful derivative analysis hinted at \(x = e^{-2}\) being a local and absolute minimum for \(f(x)\). Understanding derivatives deeply enhances your ability to explore functions thoroughly and predict their behavior accurately.