Problem 122 - Calculus Volume 1

Question

Find the local and/or absolute maxima for the functions over the specified domain. \(y=\sqrt{9-x}\) over [1,9]

Step-by-Step Solution

Answer

The absolute maximum is \(y = \sqrt{8}\) at \(x = 1\).

Step 1: Identify the Domain

The function is given as \(y = \sqrt{9-x}\) over the closed interval \([1, 9]\). This means we will find the maxima within this interval, including at the endpoints.

Step 2: Find the Derivative

To determine critical points, we first find the derivative of \(y = \sqrt{9-x}\). Using the chain rule, the derivative is \(y' = \frac{-1}{2\sqrt{9-x}}\).

Step 3: Identify Critical Points

Critical points occur where the derivative is zero or undefined. Since \(y' = \frac{-1}{2\sqrt{9-x}}\) is undefined when \(x = 9\), and never zero as \(-1 eq 0\), we need to consider \(x = 9\) as a critical point.

Step 4: Evaluate Function at Critical Points and Endpoints

Now evaluate \(y\) at the endpoints and the critical point. At \(x = 1\), \(y = \sqrt{9-1} = \sqrt{8}\). At \(x = 9\), \(y = \sqrt{9-9} = 0\).

Step 5: Determine the Maximum Value

Comparing the function values at \(x = 1\) and \(x = 9\), we find that \(y = \sqrt{8}\) (at \(x = 1\)) is greater than \(y = 0\) (at \(x = 9\)). Thus, the absolute maximum over the interval \([1, 9]\) is \(y = \sqrt{8}\) at \(x = 1\).

Key Concepts

Derivative

Derivatives are a foundational concept in calculus. They represent the rate at which a function is changing at any given point. Think of it as the slope of the tangent line to the function's graph at a particular point. For the function \( y = \sqrt{9-x} \), we use derivatives to find where the function might have its highest or lowest values. This process can help us determine the critical points, which are essential in finding local and absolute extrema. In this case, we differentiate using the chain rule, yielding the derivative \( y' = \frac{-1}{2\sqrt{9-x}} \). The negative sign indicates the function is decreasing over the interval, explaining why understanding the derivative gives insight into the function's behavior.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are crucial in calculus as they can indicate potential maxima or minima of a function. For \( y = \sqrt{9-x} \), we find our critical point by checking where the derivative \( y' = \frac{-1}{2\sqrt{9-x}} \) becomes undefined. This happens at \( x = 9 \) because the square root becomes zero, causing the denominator to be zero as well. By understanding critical points, you can identify possible places where the extrema occur, although it doesn't guarantee that these are maxima or minima without further investigation.

Absolute Maximum

An absolute maximum in a function within a given interval is the highest value that the function attains. To find it, you evaluate the function at its critical points and endpoints within the interval. For this exercise, we have the interval \([1, 9]\). The endpoints are crucial because they could be the location of the maximum if the function doesn't achieve its extrema at any other point. Upon evaluation:

• At \(x = 1\), \( y = \sqrt{8} \)
• At \(x = 9\), \( y = 0 \)

Comparing these, \( y = \sqrt{8} \) is the absolute maximum since it's greater than \( y = 0 \). So, on the interval \([1, 9]\), the function reaches its peak value at \( x = 1 \).

Chain Rule

The chain rule is a technique for finding the derivative of composite functions. A composite function is formed when one function is applied to the result of another function. In simplest terms, the chain rule helps us find how fast a layer of calculus functions within another changes.For our function \( y = \sqrt{9-x} \), think of it as: an outer function, the square root, and an inner function \( 9-x \). The chain rule states that the derivative of the composite is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This yields:

• Derivative of outer: \( \frac{1}{2\sqrt{9-x}} \)
• Derivative of inner: \( -1 \)

Thus, the derivative becomes \( y' = \frac{-1}{2\sqrt{9-x}} \). Understanding this principle allows you to tackle complex derivative problems with more ease, breaking them down into manageable parts.