Problem 28 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f \), if any, on the given interval, and state where those values occur. \(f(x)=\frac{x-2}{x+1} ;(-1,5]\)

Step-by-Step Solution

Answer

The absolute maximum value is \(\frac{1}{2}\) at \(x = 5\); no absolute minimum in \((-1,5]\).

Step 1: Identify Domain and Examine Interval Endpoints

The given function is \( f(x) = \frac{x-2}{x+1} \) and the interval is \((-1, 5]\). Notice that the function is undefined at \(x = -1\) since it makes the denominator zero. Therefore, \(-1\) is not included in the interval. We will consider \(x = 5\) which is included in the interval and check this endpoint later.

Step 2: Find the Critical Points

To find the critical points, we need to find the derivative of \( f(x) \). Using the quotient rule, the derivative is:\[ f'(x) = \frac{(x+1)(1) - (x-2)(1)}{(x+1)^2} = \frac{3}{(x+1)^2} \]The critical points occur where \( f'(x) = 0 \) or when \( f'(x) \) is undefined. Since \( \frac{3}{(x+1)^2} \) is never zero and only undefined at \( x = -1 \) (outside the domain of the function in the interval), there are no critical points in the interval.

Step 3: Evaluate the Function at the Endpoint

Since there are no critical points in the interval, we will evaluate the function \( f(x) \) at the endpoint \( x = 5 \):\[ f(5) = \frac{5 - 2}{5 + 1} = \frac{3}{6} = \frac{1}{2} \]

Step 4: Check Behavior Near \(x = -1\)

There's no need to check for absolute minima or maxima at \(x = -1\) since it's an open endpoint, but checking the behavior around it can be insightful. As \(x\) approaches \(-1\) from the right, \(f(x)\) approaches negative infinity because the denominator approaches zero from the positive side.

Step 5: Determine Absolute Maximum and Minimum Values

Considering the results, the function has an absolute minimum approaching negative infinity as \(x\) approaches \(-1\) from the right, but this isn't included within the interval as an absolute minimum since the interval is open at \(-1\). The absolute maximum value is \(\frac{1}{2}\) occurring at \(x = 5\).

Key Concepts

Absolute Maximum and Minimum

In calculus optimization, the absolute maximum and minimum values of a function are the highest and lowest values a function reaches on a given interval. When seeking these extreme values, it's crucial to examine both critical points within the interval and endpoint values.

Absolute maximum and minimum values help us understand the overall behavior of a function over a certain range. If a function reaches its absolute maximum or minimum at endpoints or critical points, these are the key values to consider. For example, in the original exercise, the absolute maximum is found at the interval endpoint \( x = 5 \), where the function evaluates to \( \frac{1}{2} \).

• Absolute Max: Highest function value in the interval
• Absolute Min: Lowest function value in the interval

Understanding absolute extrema is essential for many applications, such as finding optimum points in real-world scenarios.

Critical Points

Critical points are where the derivative of a function is either zero or undefined. Finding these points is key in determining where a function's graph might change direction or behavior, which can aid in locating potential extrema.

For the function \( f(x) = \frac{x-2}{x+1} \), the critical points come from solving when the derivative \( f'(x) \) is zero or undefined. However, in this problem's interval, the derivative \( f'(x) = \frac{3}{(x+1)^2} \) is never zero, indicating there are no critical points within the interval.

• If \( f'(x) = 0 \): Possible extremum
• If undefined: Check if outside given interval

Critical points guide us toward understanding the shape and turning behavior of a function’s graph.

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. If you have \( g(x) = \frac{u(x)}{v(x)} \), then the derivative \( g'(x) \) is calculated as:

\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]

This rule is applicable when the function you’re dealing with is expressed as a quotient. In the case of the given function, \( f(x) = \frac{x-2}{x+1} \), applying the quotient rule helped find:

\[ f'(x) = \frac{1(x+1) - (x-2)(1)}{(x+1)^2} = \frac{3}{(x+1)^2} \]

• Use where function is \( \frac{u}{v} \)
• Ensure denominator not equal to zero

The quotient rule is vital for understanding how ratios of functions behave and for simplifying complex derivative problems.

Interval Endpoints

Interval endpoints refer to the boundaries of the range of values over which you're analyzing a function. Depending on whether the interval is open or closed, these endpoints might be included or excluded.

For the given interval \((-1, 5] \), \(-1\) is an open endpoint, while \(5\) is a closed endpoint. An open endpoint, like \(-1\), is not part of the interval, so we cannot substitute it directly into the function to find an absolute extremum.

Despite this, it's wise to examine behavior as \(x\) approaches such a point. In the function \( f(x) = \frac{x-2}{x+1} \), \( f(x) \) approaches negative infinity as \( x \) approaches \(-1\) from the right.

• Closed endpoints: Include and evaluate
• Open endpoints: Exclude but check nearby behavior

Understanding interval endpoints is crucial, especially in open intervals, to correctly analyze function behavior over the specified range.