Question

We saw in Example \(3.5\) that the function $$ f(x)=\alpha \cosh (x / 4)-x $$ has two roots for \(\alpha=2\) and none for \(\alpha=10\). Is there an \(\alpha\) for which there is precisely one root? If yes, then find such an \(\alpha\) and the corresponding root; if not, then justify.

Short answer

If so, find the value of \(\alpha\). Answer: Yes, there exists a value of \(\alpha\) for which the function \(f(x) = \alpha \cosh(x/4) - x\) has exactly one root. The value of \(\alpha\) can be found using the formula: $$ \alpha = \sqrt{\frac{16}{\cosh^2\left(\frac{x}{4}\right) - 1}} $$

Step-by-step solution

Compute the derivative of the function

First, we need to compute the derivative of the given function with respect to \(x\). The function is: $$ f(x) = \alpha \cosh\left(\frac{x}{4}\right) - x $$ The derivative of the hyperbolic cosine function is the hyperbolic sine function, and the derivative of \(x\) is \(1\). Applying the chain rule for the derivative of \(\cosh(\frac{x}{4})\) with respect to \(x\), we find that: $$ f'(x) = \frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) - 1 $$

Determine where the function is increasing or decreasing

Now that we have the derivative, let's use it to determine where the function is increasing or decreasing, which will help us understand the change in the number of roots. If \(f'(x) > 0\), then the function is increasing for that value of \(x\). Similarly, if \(f'(x) < 0\), the function is decreasing. We have: $$ f'(x) = \frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) - 1 $$ Setting \(f'(x) = 0\), we get: $$ \frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) = 1 $$

Analyze the behavior of the roots with respect to \(f'(x)\) and find a value for \(\alpha\)

Since \(\cosh(x)\) and its derivative \(\sinh(x)\) are always non-negative for real numbers, the only point where \(f'(x) = 0\) must satisfy \(\frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) \geq 1\). As a result, the number of roots for the function increases from \(0\) to \(2\) as \(\alpha\) moves from a value where the inequality is satisfied to a value where it isn't. Now, for the given function to have exactly one root, there must be a point where this transition happens, i.e., where the function changes from having no roots to having two roots. This point occurs when \(\frac{\alpha}{4}\sinh\left(\frac{x}{4}\right) = 1\). So, we have: $$ \alpha \sinh\left(\frac{x}{4}\right) = 4 $$ Since we're looking for a value of \(\alpha\) such that there is exactly one root, we should note that this transition will happen at the maximum point of the function due to its shape. Therefore, we can find this maximum point by setting \(f'(x) = 0\) and solving for \(x\): $$ \frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) - 1 = 0 $$ $$ \sinh\left(\frac{x}{4}\right) = \frac{4}{\alpha} $$ Now, we can use the given condition that \(\alpha\sinh\left(\frac{x}{4}\right) = 4\): $$ \sinh\left(\frac{x}{4}\right) = \frac{4}{\alpha} $$ $$ \alpha^2\sinh^2\left(\frac{x}{4}\right) = 16 $$ We know that \(\cosh^2x - \sinh^2x = 1\), so substituting and simplifying, we have: $$ \alpha^2\left(\cosh^2\left(\frac{x}{4}\right) - 1\right) = 16 $$ $$ \alpha^2 \cosh^2\left(\frac{x}{4}\right) = 16 + \alpha^2 $$ $$ \alpha^2 (\cosh^2\left(\frac{x}{4}\right) - 1) = 16 $$ $$ \alpha^2 = \frac{16}{\cosh^2\left(\frac{x}{4}\right) - 1} $$ Now, we can find \(\alpha\): $$ \alpha = \sqrt{\frac{16}{\cosh^2\left(\frac{x}{4}\right) - 1}} $$ We know that \(\cosh^2\left(\frac{x}{4}\right) - 1 \ge 0\), so we can always find a value of \(\alpha\) which makes this equation true. Therefore, there exists an \(\alpha\) for which there is precisely one root and it is: $$ \alpha = \sqrt{\frac{16}{\cosh^2\left(\frac{x}{4}\right) - 1}} $$ The corresponding root can be found by substituting \(\alpha\) into the original function \(f(x)\) and solving for \(x\).

Key concepts

Derivative Analysis

When dealing with functions, understanding their derivative is crucial. The derivative tells us how the function behaves and helps identify where the function is increasing or decreasing. For the given function \[ f(x) = \alpha \cosh\left(\frac{x}{4}\right) - x \]we determine the derivative by using the derivative rules of hyperbolic functions. The derivative of the hyperbolic cosine function, \(\cosh(x)\), is \(\sinh(x)\). To find \(f'(x)\), we apply the chain rule, as the function involves a composition of \(\cosh\) and a linear transformation \[ f'(x) = \frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) - 1 \].Analyzing this derivative reveals where the function is increasing or decreasing. If \(f'(x) > 0\), the function increases; if \(f'(x) < 0\), it decreases. That's why we set \(f'(x) = 0\) to find "critical points," these are the points where behavior could change.

Hyperbolic Functions

Hyperbolic functions, like their trigonometric counterparts, are functions that arise in the context of hyperbolas, much like trigonometric functions arise with circles. These functions include \(\sinh(x)\) and \(\cosh(x)\). They have properties akin to sine and cosine but map out hyperbolas instead.

• Definition: The hyperbolic sine function, \(\sinh(x)\), is defined as \(\frac{e^x - e^{-x}}{2}\).
• The hyperbolic cosine function, \(\cosh(x)\), is \(\frac{e^x + e^{-x}}{2}\).

When analyzing functions, we often use these definitions to find derivatives or evaluate function behaviors. In our case, the function involves \(\cosh(x/4)\), indicating special attention to hyperbolic identities may be needed. These identities help simplify derivative calculations and solve equations. It’s important to understand how \(\sinh(x)\) and \(\cosh(x)\) interact, especially how \(\cosh^2(x) - \sinh^2(x) = 1\), which is key to resolving the solution for the roots.

Root Multiplicity Analysis

To determine the number of roots, it's crucial to find where transitions occur in the number of roots. This involves root multiplicity, which is about understanding how `simple` or `repeated` a root can be.In this context, a transition occurs where the derivative transitions from a point of zero roots to one or two roots. This is seen when analyzing \(f'(x)\) and setting the condition \[\frac{\alpha}{4} \sinh\left(\frac{x}{4}\right) = 1 \]Finding the correct \(\alpha\) ensures there's exactly one root. This scenario sets a real-world connection to root multiplicity, as we explore conditions ensuring a single tangential occurrence with the x-axis.
Identifying the precise \(\alpha\) involves solving the equation for when it just touches the x-axis. These types of analyses are crucial for understanding numerical root-finding and provide pathways to solving large systems of equations reliably.