Problem 18 - Calculus: Early Transcendentals

Question

Find the absolute extrema for \(f(x)=\ln (x) / x^{2}\) on [1,4] .

Step-by-Step Solution

Answer

The absolute minimum is 0 at \(x=1\) and the absolute maximum is \(\frac{\ln(4)}{16}\) at \(x=4\).

Step 1: Differentiation

To find the critical points, we first need to calculate the derivative of the function. Set the function as \( f(x) = \frac{\ln(x)}{x^2} \). Using the quotient rule \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \), where \( u = \ln(x) \) and \( v = x^2 \), compute the derivative. \( u' = \frac{1}{x} \) and \( v' = 2x \), so we have:\[f'(x) = \frac{(\frac{1}{x})(x^2) - (\ln(x))(2x)}{x^4} = \frac{x - 2x\ln(x)}{x^3}\]This simplifies to:\[f'(x) = \frac{1 - 2\ln(x)}{x^3}\]

Step 2: Finding Critical Points

Set the derivative equal to zero to find the critical points:\[\frac{1 - 2\ln(x)}{x^3} = 0\]This implies:\[1 - 2\ln(x) = 0\]Solving for \(x\), we get:\[2\ln(x) = 1 \\ln(x) = \frac{1}{2} \x = e^{\frac{1}{2}} = \sqrt{e}\]This gives us one critical point in the interval \([1, 4]\): \(x = \sqrt{e}\).

Step 3: Evaluate Function at Critical Points and Endpoints

Now evaluate the function \(f(x)\) at the critical point \(x = \sqrt{e}\) and at the endpoints \(x = 1\) and \(x = 4\).* \(f(1) = \frac{\ln(1)}{1^2} = 0\)* \(f(\sqrt{e}) = \frac{\ln(\sqrt{e})}{(\sqrt{e})^2} = \frac{\frac{1}{2}}{e} = \frac{1}{2e}\)* \(f(4) = \frac{\ln(4)}{4^2} = \frac{\ln(4)}{16}\).

Step 4: Determine Absolute Extrema

Compare the evaluated values to determine which are minimum and maximum.* \(f(1) = 0\)* \(f(\sqrt{e}) = \frac{1}{2e}\)* \(f(4) = \frac{\ln(4)}{16}\)Since \( \ln(4) \approx 1.386 \), \(\frac{\ln(4)}{16} \approx 0.0866\). Comparing these values: - The maximum value is \(f(4) = \frac{\ln(4)}{16}\).- The minimum value is \(f(1) = 0\).

Key Concepts

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a quotient of two functions. It's essential when dealing with functions where one variable is divided by another, as in the function from our exercise, \(f(x) = \frac{\ln(x)}{x^2}\).
To apply the quotient rule, if you have a function \(\frac{u}{v}\), the derivative is given by:

• \[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
• It requires the calculation of the derivatives of both the numerator (\(u'\)) and the denominator (\(v'\)).

For the function \(f(x)\):

• Let \(u = \ln(x)\) and \(v = x^2\).
• Thus, \(u' = \frac{1}{x}\) and \(v' = 2x\).
• The application of the quotient rule results in: \[f'(x) = \frac{(\frac{1}{x})(x^2) - (\ln(x))(2x)}{x^4} = \frac{x - 2x\ln(x)}{x^3}\]

This derivation simplifies further to the form \(f'(x) = \frac{1 - 2\ln(x)}{x^3}\). Understanding how to use the quotient rule helps in calculating derivatives of complex functions.

Critical Points

Critical points of a function are found where its derivative is zero or undefined, and they are crucial for locating extrema within a certain interval.
To find these points for a function, set its derivative to zero and solve for \(x\). In the example given:

• The derivative \(f'(x) = \frac{1 - 2\ln(x)}{x^3}\).
• Setting it to zero, we obtain: \[\frac{1 - 2\ln(x)}{x^3} = 0\]
• This simplifies to \(1 - 2\ln(x) = 0\), or \(2\ln(x) = 1\).
• Solving gives \(\ln(x) = \frac{1}{2}\), thus \(x = e^{\frac{1}{2}} = \sqrt{e}\).

In our exercise, the only critical point within the interval \([1, 4]\) is \(x = \sqrt{e}\). Evaluating the function at these critical points helps us determine potential extrema, which are points where the function might reach a maximum or minimum value.

Logarithmic Differentiation

Logarithmic differentiation is a technique used often in calculus to differentiate functions with variable exponents or products and quotients of functions. It can simplify finding derivatives for complex expressions.
It involves taking the natural logarithm of both sides of the equation and then differentiating. This method is particularly helpful when dealing with exponential functions or when the function includes roots.
For example, suppose you need to differentiate a function \(y = f(x)^{g(x)}\). Instead of the usual method, you can use logarithmic differentiation by

• Taking the logarithm of both sides: \(\ln(y) = \ln(f(x)^{g(x)}) = g(x) \ln(f(x))\).
• Then differentiate: \(\frac{1}{y} \frac{dy}{dx} = g'(x) \ln(f(x)) + g(x) \frac{f'(x)}{f(x)}\).

Returning to our problem, though we proceeded with the quotient rule, logarithmic differentiation could also simplify this process for more complex variations.
This method underscores the versatility and power of logarithms within calculus.