Problem 58 - 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. $$ f(x)=\tanh (x)-\exp (2 x) $$

Step-by-Step Solution

Answer

Find \(c\) where \(\text{sech}^2(c) = 2\exp(2c)\). Test for max/min using sign changes in \(f'(x)\).

Step 1: Find the First Derivative

To start with, differentiate the given function \(f(x) = \tanh(x) - \exp(2x)\) to find \(f'(x)\). The derivative of \(\tanh(x)\) is \(\text{sech}^2(x)\) and for \(\exp(2x)\) it's \(2\exp(2x)\). Thus, \(f'(x) = \text{sech}^2(x) - 2\exp(2x)\).

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points: \(\text{sech}^2(x) - 2\exp(2x) = 0\). Solving this equation involves finding \(x\) such that \(\text{sech}^2(x) = 2\exp(2x)\). Simplifying for \(x\) results in the critical points \(x\) that satisfy the equation.

Step 3: Apply the First Derivative Test

Evaluate the sign of \(f'(x)\) around the critical points determined in Step 2. Choose test points on either side of each \(c\) value found. If \(f'(x)\) changes from positive to negative, \(f(x)\) has a local maximum at \(x = c\). If \(f'(x)\) changes from negative to positive, \(f(x)\) has a local minimum at \(x = c\). If \(f'(x)\) does not change sign, there is neither a local maximum nor minimum at that point.

Key Concepts

Critical Points

Critical points are specific points on the graph of a function where its derivative is zero or fails to exist. In simpler terms, these are points where the function's rate of change shifts direction. To locate critical points, we differentiate the function 3), and then set the result equal to zero. This process highlights where potential peaks, valleys, or points of inflection are located.

Critical points play a vital role because they are the potential locations for local maxima and minima. Keep in mind that not all critical points represent extrema. They simply indicate potential turns in the graph's behavior. By plugging these points into the function's first derivative, you can start investigating the nature of these critical points in the larger context of the graph.

Differentiation

Differentiation is a core mathematical process that allows us to compute a function's derivative. The derivative, denoted as \(f'(x)\), describes how quickly the function's value changes as \(x\) changes. In our exercise, we need this to investigate the behavior of the function \(f(x)=\tanh (x) - \exp(2x)\).

To differentiate \(f(x)\), apply the rules of differentiation separately to each term. In this case, for \(\tanh(x)\), the derivative is \(\text{sech}^2(x)\), and for \(\exp(2x)\), it is \(2\exp(2x)\). Combining these, we have \(f'(x)= \text{sech}^2(x) - 2\exp(2x)\).

With the derivative in hand, you have a tool that tells you not only where the critical points are, but also how the function behaves around those points.

Local Extrema

Local extrema are points at which the function reaches its greatest or smallest value relative to the points immediately surrounding it. Determining whether a critical point is a local maximum, a local minimum, or neither involves evaluating the first derivative at points around the critical point.

To begin, use the First Derivative Test. Here1the critical point \(c\) obtained from setting \(f'(x) = 0\) must be analyzed by choosing points slightly less than and greater than \(c\).

• If \(f'(x)\) changes from positive to negative around \(c\), the function has a local maximum there.
• If \(f'(x)\) switches from negative to positive, then it's a local minimum.
• If there's no sign change, then no local extreme occurs at that point.

Understanding local extrema is crucial for sketching the behavior of a function's graph and predicting how it operates in its domain.