Problem 31 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The relative extrema for this function can be found at \(x = 1\) and \(x = e\) if they pass the Second Derivative Test.

Step 1: Find the derivative

The derivative of \(y=\frac{x}{\ln x}\) can be found using the quotient rule \( (u/v)' = (vu' - uv')/v^2 \). Treat \(x\) as \(u\) and \(\ln x\) as \(v\). Thus, \(u'=1\) and \(v'=1/x\). Implementing these into the formula gives \(y' = \frac{\ln x \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}\).

Step 2: Find critical numbers

The critical numbers occur where the derivative is zero or undefined. Thus, set \(y' = \frac{\ln x - 1}{(\ln x)^2} = 0\) and solve for \(x\). That gives \(\ln x = 1\), which implies \(x = e\). To find numbers where \(y'\) is undefined, set the denominator equal to zero and solve for \(x\): \((\ln x)^2 = 0\), which gives \(x = 1\). Therefore, the critical numbers are \(x = 1\) and \(x = e\).

Step 3: Second Derivative Test

Take the second derivative of \(y\): \(y''=(y')' = (\frac{\ln x - 1}{(\ln x)^2})' \). Once you have the second derivative, plug in the critical numbers. If \(y''(1)\) or \(y''(e) > 0\), then \(x = 1\) or \(x = e\) is a local minimum. If it is <0, then they are local maximums, and if \(y''(1)\) or \(y''(e) = 0\), the Second Derivative Test is inconclusive. In this case, one would have to apply the First Derivative Test to determine the nature of the critical number.

Key Concepts

Relative Extrema

Finding relative extrema is a crucial part of calculus problem solving. These extrema, also known as local maxima or minima, indicate the highest or lowest points in a certain interval of a function. To identify these points, one often needs to find where the derivative of a function is equal to zero or does not exist. In simple terms, when the slope of the tangent line is zero or undefined, it signals a potential extremum.

For the function \( y = \frac{x}{\ln x} \), these relative extrema manifest at critical numbers, which we find by solving the derivative expression for cases where it equals zero or is undefined. In this case, we found that critical numbers are at \( x = 1 \) and \( x = e \). It is important to note that critical numbers don't automatically mean an extremum exists; further verification with tests like the second derivative test is needed.

Quotient Rule

The quotient rule is a handy differentiation technique used when dealing with ratios of two functions. If you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then its derivative can be found as:

• \( (\frac{u}{v})' = \frac{v u' - u v'}{v^2} \)

In this formula, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively. Using the quotient rule for \( y = \frac{x}{\ln x} \), we set \( u = x \) and \( v = \ln x \). By substituting \( u' = 1 \) and \( v' = \frac{1}{x} \), we derive...

• \( y' = \frac{\ln x \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2} \)

The quotient rule thus helps break down complex fractions into manageable derivatives, facilitating the process of finding extrema or optimizing functions.

Second Derivative Test

The second derivative test provides a sophisticated method to determine the concavity of a function at its critical points, helping decide whether these points are local minima or maxima. Once critical numbers are found, the second derivative \( y'' \) is calculated, then evaluated at these critical numbers.

For \( y = \frac{x}{\ln x} \), having the first derivative \( y' = \frac{\ln x - 1}{(\ln x)^2} \), you take its derivative again to get \( y'' \), and plug in each critical number:

• If \( y''(x) > 0 \), the function is concave up at \( x \), indicating a local minimum.
• If \( y''(x) < 0 \), the function is concave down, which points to a local maximum.
• If \( y''(x) = 0 \), the test is inconclusive, and other tests like the first derivative test might be needed.

Using the second derivative test complements the process initiated by finding derivatives, assisting in painting a complete picture of the function’s behavior at its critical numbers.