Problem 98 - Calculus Single Variable

Question

In each of Exercises \(91-94\), calculate and plot the derivative \(f^{\prime}\) of the given function \(f\). Use this plot to identify candidates for the local extrema of \(f\). Add the plot of \(f\) to the window containing the graph of \(f^{\prime} .\) From this second plot, determine the behavior of \(f\) at each candidate for a local extremum. $$ f(x)=x \sin (1 / x), \quad I=[1 /(4 \pi), \pi / 24] $$

Step-by-Step Solution

Answer

Find critical points of \( f'(x) \), plot both \( f(x) \) and \( f'(x) \) together, assess local extrema behavior by observing changes in \( f'(x) \).

Step 1: Find the Derivative

To calculate the derivative of the given function \( f(x) = x \sin\left(\frac{1}{x}\right) \), we use the product and chain rules. First, identify \( u(x) = x \) and \( v(x) = \sin\left(\frac{1}{x}\right) \). The derivative using the product rule is \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Here, \( u'(x) = 1 \) and \( v'(x) = \cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right) \). So, \( f'(x) = 1 \cdot \sin\left(\frac{1}{x}\right) + x \cdot \left(-\frac{1}{x^2}\right)\cos\left(\frac{1}{x}\right) \). Simplify this to \( f'(x) = \sin\left(\frac{1}{x}\right) - \frac{1}{x} \cos\left(\frac{1}{x}\right) \).

Step 2: Plot the Derivative

Using a graphing tool or calculator, plot \( f'(x) \) over the interval \( I = \left[ \frac{1}{4\pi}, \frac{\pi}{24} \right] \). This will help in identifying critical points where the derivative could be zero, indicating potential local extrema.

Step 3: Identify Critical Points

Examine the plot of \( f'(x) \). Look for points where the derivative crosses the x-axis (i.e., \( f'(x) = 0 \)) or where \( f'(x) \) does not exist, as these points can correspond to local extrema of \( f(x) \).

Step 4: Behavior at Critical Points

For each critical point identified in Step 3, analyze the surrounding intervals to determine the behavior of \( f(x) \). If \( f'(x) \) changes from positive to negative, the point is a local maximum. If \( f'(x) \) changes from negative to positive, it is a local minimum.

Step 5: Plot the Original Function

Plot \( f(x) = x \sin\left(\frac{1}{x}\right) \) over the same interval \( I \) in the same window as the plot of \( f'(x) \). This helps in visually confirming the behavior of \( f(x) \) at the identified critical points.

Key Concepts

Product Rule

When differentiating a function that is the product of two separate functions, such as the function given in the exercise \( f(x) = x \sin\left(\frac{1}{x}\right) \), the product rule becomes essential. The product rule provides a systematic way to find the derivative of a product, ensuring accuracy and efficiency. The rule is defined as follows:

• If you have two functions, \( u(x) \) and \( v(x) \), then the derivative \( (uv)' \) can be calculated by: \( u'(x)v(x) + u(x)v'(x) \).
• In our exercise, we identified \( u(x) = x \) and \( v(x) = \sin\left(\frac{1}{x}\right) \).

To find \( f'(x) \), we compute:

• \( u'(x) = 1 \), since the derivative of \( x \) is 1.
• Combine this with \( \frac{d}{dx}\sin\left(\frac{1}{x}\right) \) for \( v'(x) \), which we'll explore more in the chain rule section.

Use the product rule formula:

• \( f'(x) = 1 \cdot \sin\left(\frac{1}{x}\right) + x \cdot v'(x) \).

Chain Rule

The chain rule is indispensable when dealing with compositions of functions, especially those involving trigonometric or complex powers. In our function, \( v(x) = \sin\left(\frac{1}{x}\right) \), we encounter such a situation.To differentiate \( \sin\left(\frac{1}{x}\right) \), you treat \( \frac{1}{x} \) as an inner function and \( \sin \) as the outer function:

• The derivative of \( \sin(u) \), where \( u = \frac{1}{x} \), is \( \cos(u) \cdot u'(x) \).
• Calculate \( u'(x) = \left( \frac{1}{x} \right)' = -\frac{1}{x^2} \).
• Thus, \( v'(x) = \cos\left(\frac{1}{x}\right) \cdot \left( -\frac{1}{x^2} \right) \).

Combining the two rules:

• Substitute \( v'(x) \) back into the product rule to get the full derivative expression: \( f'(x) = \sin\left(\frac{1}{x}\right) - \frac{1}{x} \cos\left(\frac{1}{x}\right) \).

Local Extrema

Local extrema of a function are the points where a function reaches a local maximum or minimum. To identify these points, we look for where the derivative is zero or undefined. These critical points are potential candidates for local extrema.When the derivative changes sign (from positive to negative or vice-versa), it indicates a local extremum:

• A sign change from positive to negative in \( f'(x) \) signifies a local maximum.
• A change from negative to positive indicates a local minimum.

Analyzing the plot of \( f'(x) \), you'll notice that whenever the curve crosses the x-axis, these are points of interest:

• Further analyze the critical points on the graph of \( f(x) \) visualized together with its derivative, which aids in confirming the type of extremum and ensures a comprehensive understanding.
• By examining the intervals around these critical points, you fully determine the nature of each extremum.