Problem 20 - Calculus Single Variable

Question

In each of Exercises \(7-22,\) use Fermat's Theorem to locate each \(c\) for which \(f(c)\) is a candidate extreme value of the given function \(f\) $$ f(x)=\sqrt{x}+2 / \sqrt{x} $$

Step-by-Step Solution

Answer

The potential extreme value candidate is \( x = 2 \).

Step 1: Understand Fermat's Theorem

Fermat's Theorem states that if a function has a local extremum (either a minimum or maximum) at some point in its domain and the derivative exists at that point, then the derivative of the function is zero at that point.

Step 2: Calculate the Derivative of f(x)

To apply Fermat's Theorem, find the derivative of the given function, \[ f(x) = \sqrt{x} + \frac{2}{\sqrt{x}} \].First, rewrite the function as \[ f(x) = x^{1/2} + 2x^{-1/2} \]. Then, the derivative is \[ f'(x) = \frac{1}{2}x^{-1/2} - x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{x\sqrt{x}} \].

Step 3: Set the Derivative Equal to Zero

Using Fermat's Theorem, find the values of \( x \) where the derivative is zero:\[ \frac{1}{2\sqrt{x}} - \frac{1}{x\sqrt{x}} = 0 \].Multiply through by \( x\sqrt{x} \) to simplify:\[ \frac{x}{2} - 1 = 0 \].

Step 4: Solve for x

Solve the equation \( \frac{x}{2} - 1 = 0 \) to find the critical points:\[ \frac{x}{2} = 1 \],\[ x = 2 \].

Step 5: Verify if x = 2 is in the Domain

Check if \( x = 2 \) is in the domain of \( f(x) \), which is \( x > 0 \). Since \( x = 2 \) is positive and in the domain, it is a candidate for extreme values.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with continuous change. It primarily involves two fundamental concepts: differentiation and integration. Differentiation deals with finding the rate at which a quantity changes, while integration involves finding the total accumulation of a quantity. These concepts allow us to analyze functions and their properties, such as growth rates, areas under curves, and extremum points.

In the context of the given exercise, we are interested in the differentiation aspect of calculus. We use differentiation to find critical points, which helps us in identifying extremum (maximum or minimum) values of the function. This process involves computing the derivative of the function, and using tools such as Fermat’s Theorem to locate these extremum values efficiently. Thus, calculus provides the mathematical framework and techniques needed to solve such problems.

Calculus makes intricate relationships between quantities manageable and presents a powerful way to model and solve real-world problems, ranging from physics to economics.

Critical Points

Critical points of a function are points in its domain where the derivative is either zero or undefined. These points are important because they represent potential locations for a function’s local maxima or minima. When analyzing a function, finding these points is crucial to understanding its behavior.

To find the critical points in the exercise, we first calculate the derivative of the function. Then we set this derivative equal to zero, because according to Fermat's Theorem, local maxima and minima occur where the derivative is zero, provided it exists at that point. However, not all critical points will necessarily be extremum points; they need to be further analyzed to determine their nature.

In our example, after finding the derivative, we solve the equation \( \frac{x}{2} - 1 = 0 \) to obtain \( x = 2 \). Since this value lies in the domain of the function, it qualifies as a critical point which might be an extremum.

Derivative

A derivative represents the rate of change of a function with respect to one of its variables. It provides a precise way to study how a function behaves as its inputs vary. In essence, if you think of a function as describing a mountain landscape, the derivative tells you how steeply you’re climbing up or down at any given point.

For the function \( f(x) = \sqrt{x} + \frac{2}{\sqrt{x}} \), finding the derivative involves applying rules of calculus to express how small changes in \( x \) affect \( f(x) \). The derivative is given by \( f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x\sqrt{x}} \).

From the derivative, we set it to zero to solve for the values of \( x \) where the slope is zero, indicating potential critical points. This step is crucial as it forms the basis for determining where the function achieves its relative maximums or minimums, helping to solve the problem of finding the extremum values.