Problem 55 - Algebra and Trigonometry

Question

\(51-58=\) Find the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. $$ U(x)=x \sqrt{6-x} $$

Step-by-Step Solution

Answer

The local maximum is approximately 5.66 at x = 4; there's no distinct local minimum.

Step 1: Find the Derivative

To find the local maxima and minima of the function, we need to find its derivative. The given function is \( U(x) = x \sqrt{6-x} \). Using the product rule and the chain rule, the derivative is: \[ U'(x) = \frac{d}{dx}(x) \cdot \sqrt{6-x} + x \cdot \frac{d}{dx}(\sqrt{6-x}) \].Let's calculate each part:- Derivative of \( x \) is 1.- Derivative of \( \sqrt{6-x} \) is \( \frac{-1}{2\sqrt{6-x}} \).So, \( U'(x) = \sqrt{6-x} + x \cdot \frac{-1}{2\sqrt{6-x}} = \sqrt{6-x} - \frac{x}{2\sqrt{6-x}} \). This can be simplified by finding a common denominator: \( \frac{2(6-x) - x}{2\sqrt{6-x}} = \frac{12 - 3x}{2\sqrt{6-x}} \).

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative to zero: \( \frac{12 - 3x}{2\sqrt{6-x}} = 0 \). Since the denominator \( 2\sqrt{6-x} \) cannot be zero within the domain, set the numerator equal to zero: \( 12 - 3x = 0 \).Solving for \( x \), we get \( x = 4 \).

Step 3: Determine the Domain

Next, verify the domain of the function \( U(x) = x \sqrt{6-x} \), where \( \sqrt{6-x} \) implies that \( 6-x \geq 0 \), or \( x \leq 6 \). Since the square root must be non-negative and \( x \) also has to be non-negative, the practical domain is \( 0 \leq x \leq 6 \).

Step 4: Evaluate the Endpoints and Critical Points

Calculate the function value at the critical point and the endpoints of the domain.- At \( x = 0 \): \( U(0) = 0 \times \sqrt{6-0} = 0 \).- At \( x = 4 \): \( U(4) = 4 \sqrt{6-4} = 4\sqrt{2} \approx 5.66 \).- At \( x = 6 \): \( U(6) = 6 \times \sqrt{6-6} = 0 \).

Step 5: Identify Local Maximum and Minimum Points

Comparing the function values:- \( U(0) = 0 \), \( U(4) \approx 5.66 \), \( U(6) = 0 \).The function has its maximum value of approximately 5.66 at \( x = 4 \) and endpoints have the minimum value 0.

Key Concepts

Understanding the Derivative

Finding the derivative of a function is crucial in the study of calculus. It measures how a function changes as its input changes. You can think of it as the function's "slope" at any given point.

In our exercise, we are finding the derivative of the function:

• Given function: \( U(x) = x \sqrt{6-x} \)

To achieve this, both the product and chain rules are applied.

• Product Rule: Used when differentiating products of two functions.
• Chain Rule: Used when differentiating compositions of functions.

By applying these rules, we calculated the derivative to be:\[ U'(x) = \frac{12 - 3x}{2\sqrt{6-x}} \]

This formula shows how the function \( U(x) \) changes with \( x \). Understanding derivatives helps in identifying where functions may have peaks or valleys.

Identifying Critical Points

Critical points are where the derivative is zero or undefined. These are key points in determining the local maxima or minima of a function.

To find critical points:

• Set the derivative \( U'(x) \) to zero: \( \frac{12 - 3x}{2\sqrt{6-x}} = 0 \)
• Focus on the numerator, as the denominator must stay non-zero within the domain.
• Solve \( 12 - 3x = 0 \) to find \( x = 4 \).

Since the original function's domain is \( 0 \leq x \leq 6 \), our critical point \( x = 4 \) is valid and falls within this range. This process makes sure we catch all possible spots where the function might reach its peaks or troughs.

Function Evaluation at Critical and Endpoint Values

Evaluating a function at specific points can tell us a lot about its behavior. In particular, we look at critical points and endpoints of a function's domain to find local maxima and minima.

For the function \( U(x) = x \sqrt{6-x} \):

• Critical point: \( x = 4 \). Evaluating the function here gives \( U(4) = 4\sqrt{2} \approx 5.66 \).
• Endpoints: \( x = 0 \) and \( x = 6 \), where \( U(0) = 0 \) and \( U(6) = 0 \).

By comparing these values:

• The function's maximum value is approximately \( 5.66 \) at \( x = 4 \).
• The minimum value is \( 0 \) at the endpoints \( x = 0 \) and \( x = 6 \).

Thus, function evaluation at critical and endpoint values is key to understanding the overall behavior of the function.

Introduction to Calculus

Calculus is a branch of mathematics focused on the study of change, involving derivatives and integrals. Its concepts are used to solve problems involving dynamic systems.

• Calculus allows us to model and solve real-world problems, like predicting weather changes, optimizing systems, and understanding natural phenomena.
• Central to calculus are derivatives, which tell us about the rate of change, and integrals, which help us find accumulated quantities.

In the context of our exercise, we used calculus to:

• Determine the derivative of a function to identify critical points.
• Evaluate these points to find local maxima and minima.

By understanding calculus concepts, students can analyze and infer the behavior of more complex functions effectively.