Problem 40 - Calculus of a Single Variable

Question

In Exercises 37–40, find any relative extrema of the function. Use a graphing utility to confirm your result. $$ h(x)=2 \tanh x-x $$

Step-by-Step Solution

Answer

There are two relative minimum points at \( \pm 0.88 \).

Step 1: Find the derivative

First, take the derivative of the function. The derivative of \( \tanh x \) is \( \sech^{2}x \) and the derivative of \( x \) is 1. Therefore, \( h'(x) = 2 \sech^{2}x - 1 \).

Step 2: Identify critical points

Critical points are found where the first derivative equals zero or is undefined. Equation \( 2 \sech^{2}x - 1 = 0 \) is simplified to \( \sech^{2}x = 0.5 \) and then to \( \cosh^{2}x = 2 \). The hyperbolic cosine function is always more than or equal to 1, thus for the square to be 2, x values must be \( \pm \cosh^{-1}{\sqrt{2}} \), approximately \( \pm 0.88 \).

Step 3: Apply Second Derivative Test

To classify these critical points, apply the second derivative test. Compute the second derivative, \( h''(x) = -4 \sech^{2}x \tanh x \). Substituting the critical points, both give positive values, it indicates the critical points are relative minimums.

Key Concepts

Critical Points

Understanding critical points is fundamental when analyzing functions and their extrema. Critical points occur when the first derivative of a function is either equal to zero or is undefined. They are essential because they help identify where the graph of a function may have a horizontal tangent line or a cusp, which are the usual suspects for local maxima or minima, or other interesting behavior.

For instance, let's consider the function provided: \( h(x)=2 \tanh x-x \). To find its critical points, we take its derivative and look for where this derivative equals zero. So when we calculate \( h'(x) = 2 \sech^{2}x - 1 \) and set it equal to zero, we're essentially finding the x-values where the slope of \( h(x) \) is horizontal. After solving \( 2 \sech^{2}x - 1 = 0 \), we get the critical points as approximately \( \pm 0.88 \), which are essential to determining the behavior of the function on the graph.

Derivative of Hyperbolic Functions

Hyperbolic functions, such as \( \tanh x \) and \( \sech x \), behave similarly to trigonometric functions and are prevalent in certain areas of calculus, particularly when dealing with problems that involve rates of change in a hyperbolic context. The derivatives of hyperbolic functions help us find critical points where the function's graph changes direction.

For example, in the derivative of \( h(x) = 2 \tanh x - x \), we see the term \( 2 \sech^{2}x \) arise, which is the derivative of \( 2 \tanh x \). It is crucial to understand how the basic derivatives of hyperbolic functions are derived and how to apply them. Remember that the function \( \tanh x \) is expressed as the ratio of \( \sinh x \) to \( \cosh x \), and the derivative of \( \tanh x \) results in \( \sech^{2}x \), which helps in finding the critical points for the function.

Second Derivative Test

The second derivative test is a convenient method to determine the nature of the critical points found. By taking the second derivative of a function and evaluating it at the critical points, we can classify these points as either relative maxima, minima, or points of inflection. If the second derivative at a critical point is positive, the function has a local minimum there; if it's negative, the function has a local maximum. If it's zero, the test is inconclusive, and one may need to explore further methods such as the first derivative test or higher-order derivatives.

For our practice function \( h(x) \), after finding the second derivative \( h''(x) = -4 \sech^{2}x \tanh x \), we substitute the critical points to test each. The positive result implies that the function \( h(x) \) has relative minimums at these points. This is a powerful way to not only find but also classify the extrema of the function.