Problem 60 - Calculus Single Variable

Question

In each of Exercises 54-60, determine each point \(c,\) where the given function \(f\) satisfies \(f^{\prime}(c)=0\). At each such point, use the First Derivative Test to determine whether \(f\) has a local maximum, a local minimum, or neither. $$ f(x)=\tanh ^{-1}(\sqrt{x})-2 \sqrt{x} $$

Step-by-Step Solution

Answer

The function has a local maximum at \( x = \frac{1}{2} \).

Step 1: Verify the Domain of f(x)

The function given is \( f(x)=\tanh ^{-1}(\sqrt{x})-2 \sqrt{x} \). To ensure that \( f(x) \) is defined, \( \sqrt{x} \) must be between -1 and 1 because the domain of \( \tanh^{-1}(x) \) is \(-1 < x < 1\). Since \( \sqrt{x} \) is only defined for \( x \geq 0 \), the domain of \( f(x) \) is \( 0 \leq x < 1 \).

Step 2: Differentiate f(x) to find f'(x)

We need to find the derivative of \( f(x) \), which is \( f'(x) \). Using the chain rule and the derivative of \( \tanh^{-1}(x) \), the derivative is:\[ f'(x) = \frac{d}{dx}\left(\tanh^{-1}(\sqrt{x})\right) - \frac{d}{dx}(2\sqrt{x}) \]This simplifies to:\[ f'(x) = \frac{1}{1-x}\cdot\frac{1}{2\sqrt{x}} - \frac{2}{2\sqrt{x}} \]Simplifying further gives:\[ f'(x) = \frac{1}{2\sqrt{x}(1-x)} - \frac{1}{\sqrt{x}} \]\[ f'(x) = \frac{1 - 2x}{2x(1-x)} \]

Step 3: Find Critical Points by Solving f'(x) = 0

Set \( f'(x) = 0 \) to find the critical points:\[ \frac{1 - 2x}{2x(1-x)} = 0 \]The numerator gives us the equation \( 1 - 2x = 0 \). Solving this:\[ 1 = 2x \]\[ x = \frac{1}{2} \]Thus, \( x = \frac{1}{2} \) is the critical point.

Step 4: Use First Derivative Test for Local Extrema

We apply the First Derivative Test around \( x = \frac{1}{2} \):- For \( x < \frac{1}{2} \), choose a test point like \( x = \frac{1}{4} \): - \( f'(\frac{1}{4}) = \frac{1 - 2(\frac{1}{4})}{2(\frac{1}{4})(1-\frac{1}{4})} = \frac{1 - \frac{1}{2}}{\frac{1}{8} \cdot \frac{3}{4}} > 0 \), so \( f'(x) > 0 \).- For \( x > \frac{1}{2} \), choose a test point like \( x = \frac{3}{4} \): - \( f'(\frac{3}{4}) = \frac{1 - 2(\frac{3}{4})}{2(\frac{3}{4})(1-\frac{3}{4})} = \frac{1 - \frac{3}{2}}{\frac{3}{8} \cdot \frac{1}{4}} < 0 \), so \( f'(x) < 0 \).Since \( f'(x) \) changes from positive to negative at \( x = \frac{1}{2} \), \( f(x) \) has a local maximum at \( x = \frac{1}{2} \).

Key Concepts

First Derivative Test

The First Derivative Test is an essential method to find local extrema (maxima or minima) of a function. Once we find the critical points where the derivative equals zero or is undefined, the First Derivative Test helps us analyze the nature of these points.

To apply the test, follow these steps:

• Pick a test point on either side of each critical point.
• Evaluate the sign of the derivative at these test points.
• Observe how the signs change as you pass through the critical point.

If the derivative changes from positive to negative, the point is a local maximum. If it transitions from negative to positive, it's a local minimum. If there's no change, the point could be a point of inflection but not an extremum. The test is both simple and effective, enhancing understanding & analysis of function behavior.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points could indicate where the function changes its increasing or decreasing behavior, hence possibly marking locations for local maxima or minima.

To find critical points:

• Calculate the derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Check anywhere the derivative doesn't exist.

Critical points give us significant clues into the function's graph shape. In the analysis of the function provided, the critical point at \(x = \frac{1}{2}\) helped determine areas of rising and falling slopes, ultimately aiding in local extremum identification.

Local Maximum

A local maximum is a point where a function reaches a peak within a certain interval, but not necessarily globally. Essentially, it is where the function's value stops increasing and starts decreasing within a neighborhood around that point.

We find it often:

• By identifying a critical point where the First Derivative Test shows the derivative changing from positive to negative.

Understanding local maxima is vital for interpreting graphs as they might indicate optimal points for processes modeled by the function. In the exercise, the First Derivative Test showed a transition from positive to negative at \(x = \frac{1}{2}\), confirming a local maximum at that point.

Derivative

A derivative tells us the rate of change or the slope of a function at any given point. It indicates how a function behaves as its input changes, elucidating trends like rising or falling behavior.

To calculate a derivative, you might need rules like the chain rule, product rule, or quotient rule depending on the function's complexity.

• The derivative is zero at critical points where potential extrema occur.
• Knowing how to differentiate functions accurately is crucial for effective calculus application.

In the provided example, differentiating involved understanding the derivative of \(\tanh^{-1}(x)\) and chain rules, leading to insights on the function's behavior and its local extrema analysis.