Problem 30 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(y=x \ln x\)

Step-by-Step Solution

Answer

The function \(y = x \ln x\) has a local minimum at \(x = \frac{1}{e}\).

Step 1: Compute the first derivative

The first derivative of \(y = x \ln x\) can be found using the product rule for differentiation, which states that the derivative of two multiplied functions is the first function times the derivative of the second plus the second function times the derivative of the first. So, the derivative \(\(y' = x'\ln x + x(\ln x)'\) = \(\ln x + 1\).

Step 2: Find the critical points

Critical points exist at x-values where the first derivative is zero or does not exist. Solving \(\ln x + 1 = 0\) for \(x\), you find that the only critical point is \(x = \frac{1}{e}\).

Step 3: Compute the second derivative

The second derivative is the derivative of the first derivative. In this case, the derivative of \(\ln x + 1\) is \(\frac{1}{x}\).

Step 4: Apply the Second Derivative Test

Testing \(x = \frac{1}{e}\) in the second derivative (i.e. evaluating \(\frac{1}{x}\) at \(x = \frac{1}{e}\)), you will get \(e\), which is greater than 0. Hence, by the Second Derivative Test, \(x = \frac{1}{e}\) is a local minimum.

Key Concepts

Relative Extrema

In calculus, finding relative extrema of a function means identifying its peak or valley points within a specific interval. A relative extremum can either be a relative maximum or a relative minimum. These are points where the function changes direction, from increasing to decreasing or vice versa.

• Relative Maximum: A point where the function has the greatest value relative to its nearby values.
• Relative Minimum: A point where the function has the smallest value relative to its nearby values.

For the function \(y = x \ln x\), we're interested in where these extrema are located. To locate these points, we utilize derivatives and tests such as the First Derivative Test and the Second Derivative Test. This ensures we understand how the function behaves at different points.

First Derivative

The first derivative represents the slope of the tangent to the curve at any given point on the function. It provides information on the rate of change of the function.
To find the first derivative of \(y = x \ln x\), we apply the product rule because the function is a product of two terms: \(x\) and \(\ln x\). The product rule is given by \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).
For \(y = x \ln x\):

• First Term: \( u = x \), deriving this gives \( u' = 1 \).
• Second Term: \( v = \ln x \), deriving this gives \( v' = \frac{1}{x} \).

Putting these into the product rule, we get the first derivative \(y' = \ln x + 1\). Finding where this equals zero or doesn't exist helps identify potential extremum points.

Second Derivative Test

The second derivative test is an efficient way to determine whether a particular critical point is a relative extremum or not. After computing the first derivative, subsequent calculations of its derivative give us the second derivative. This provides information about the concavity of the function.
The second derivative tells us:

• If it is positive at a critical point, the function is concave upward at that point, indicating a relative minimum.
• If it is negative, the function is concave downward, indicating a relative maximum.
• If it is zero, the test is inconclusive.

Examining the function \(y = x \ln x\), the second derivative is \( \frac{1}{x} \). Evaluating this at the critical point \( x = \frac{1}{e} \) gives \( e \), which confirms that this point is a relative minimum due to the positive second derivative.

Critical Points

Critical points occur where the first derivative is zero or undefined, indicating possible locations for relative extrema. Identifying these points is a key step in analyzing the behavior of a function.
For the function \( y = x \ln x \), the critical points are found by setting the first derivative to zero: \( \ln x + 1 = 0 \). Solving this equation shows that the critical point is \( x = \frac{1}{e} \).

• Critical points are not guaranteed to be extrema. They are points of interest where the function's behavior changes.
• Further tests, such as the second derivative test, are required to classify them as relative maxima, minima, or neither.

Understanding critical points helps us better interpret how the function evolves and changes across its domain, identifying any high or low points.