Problem 19 - Calculus

Question

Find the derivatives of the functions \(9-18:\) Find the minimum value of \(\cosh (\ln x)\) for \(x>0\)

Step-by-Step Solution

Answer

The minimum value is 1 at \(x = 1\).

Step 1: Express Hyperbolic Cosine in Terms of Exponential Function

First, recall the definition of the hyperbolic cosine: \[\cosh(x) = \frac{e^x + e^{-x}}{2}\]So, for \(\cosh(\ln x)\), substitute \(x\) with \(\ln x\): \[\cosh(\ln x) = \frac{e^{\ln x} + e^{-\ln x}}{2} = \frac{x + \frac{1}{x}}{2}\]

Step 2: Simplify the Function

The function to minimize now becomes \(f(x) = \frac{x + \frac{1}{x}}{2}\). This step simplifies \(\cosh(\ln x)\) in terms of \(x\).

Step 3: Find the Derivative

Find the derivative of \(f(x) = \frac{x + \frac{1}{x}}{2}\):\[f'(x) = \frac{1}{2} \left(1 - \frac{1}{x^2}\right)\]

Step 4: Set the Derivative to Zero to Find Critical Points

To find the critical points, set the derivative equal to zero:\[1 - \frac{1}{x^2} = 0\]Solve for \(x\):\[x^2 = 1 \quad \Rightarrow \quad x = 1\]Since \(x > 0\), we only consider \(x = 1\).

Step 5: Determine Concavity to Confirm Minimum

Evaluate the second derivative, \(f''(x)\), to determine concavity:\[f''(x) = \frac{1}{2} \left(\frac{2}{x^3}\right) = \frac{1}{x^3}\]Since \(f''(1) = \frac{1}{1^3} = 1\) which is positive, the function is concave up at \(x = 1\), confirming a minimum.

Step 6: Evaluate Minimum Value

Now evaluate \(f(x)\) at \(x = 1\) to find the minimum value:\[f(1) = \frac{1 + \frac{1}{1}}{2} = 1\]

Key Concepts

Hyperbolic Functions

Let's start by understanding what hyperbolic functions are. They are analogs of the trigonometric functions, but they are based on hyperbolas instead of circles. Similar to how sine and cosine are related to the unit circle, hyperbolic sine and cosine are related to the hyperbola.The hyperbolic cosine function, denoted as \(\cosh(x)\), is defined as:\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]This function is essential because it appears frequently in calculus, especially in scenarios involving exponential growth and decay, as well as certain types of calculus problems.In our particular problem, we substitute \(x\) with \(\ln x\) in \(\cosh(x)\) to get \(\cosh(\ln x)\). After substituting and simplifying, this expression becomes \(\frac{x + \frac{1}{x}}{2}\), which we can then work with using calculus methods.

Derivatives

A derivative measures how a function changes as its input changes—essentially, it represents the slope of the function's graph. If you have a function \(f(x)\), the derivative \(f'(x)\) tells us how \(f(x)\) changes with respect to \(x\).For our scenario, the function \(f(x) = \frac{x + \frac{1}{x}}{2}\) needs to be derived. The first derivative, \(f'(x)\), is calculated by differentiating each component:\[ f'(x) = \frac{1}{2} \left(1 - \frac{1}{x^2}\right) \]This expression helps us locate where the function reaches its critical points, where the slope is zero or undefined. These points are crucial for understanding the behavior of the function, such as finding its minima or maxima.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points could represent the function's local maxima, minima, or points of inflection, but further analysis is needed to know for sure.In our problem, the derivative \(f'(x) = \frac{1}{2} \left(1 - \frac{1}{x^2}\right)\) is set to zero to find critical points:\[ 1 - \frac{1}{x^2} = 0 \]Solving this equation gives us \(x^2 = 1\), which results in \(x = 1\) since we're only considering positive \(x\) values due to the domain restrictions.Identifying this critical point allows us to further analyze the function using additional tools, such as the second derivative test, to confirm if it corresponds to a minimum or maximum.

Second Derivative Test

The second derivative test helps determine the concavity of a function at its critical points. If the second derivative \(f''(x)\) is positive, the function is concave up, indicating a local minimum at that critical point.For our function \(f(x) = \frac{x + \frac{1}{x}}{2}\), the second derivative becomes:\[ f''(x) = \frac{1}{x^3} \]Evaluating this at the critical point \(x = 1\), we find:\[ f''(1) = \frac{1}{1^3} = 1 \]Since \(f''(1)\) is positive, the function is concave up at \(x = 1\), confirming a local minimum. Hence, \(x = 1\) is indeed the point where \(\cosh(\ln x)\) attains its minimum value, which is \(1\). This test provides a clear indication that our function achieves a minimum at the identified point, validating our solution.