Problem 23 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ f(x)=\frac{x^{2}-3}{x-2}, \quad x \neq 2 $$

Step-by-Step Solution

Answer

Increasing on \((-\infty, 1)\) and \((3, \infty)\); local max at \((1, 2)\); local min at \((3, 6)\).

Step 1: Find the Derivative

\(f(x) = \frac{x^2 - 3}{x - 2}\)
Using the quotient rule:
\(f'(x) = \frac{2x(x-2) - (x^2-3)(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2 + 3}{(x-2)^2} = \frac{x^2 - 4x + 3}{(x-2)^2} = \frac{(x-1)(x-3)}{(x-2)^2}\)

Step 2: Find Critical Points

\(f'(x) = 0\) when \(x = 1\) or \(x = 3\). \(f'\) is undefined at \(x = 2\).

Step 3: Determine Intervals

\(f'(x) > 0\) on \((-\infty, 1)\) and \((3, \infty)\) (increasing).
\(f'(x) < 0\) on \((1, 2)\) and \((2, 3)\) (decreasing).

Step 4: Identify Extrema

Local max at \(x = 1\): \(f(1) = \frac{1-3}{1-2} = \frac{-2}{-1} = 2\).
Local min at \(x = 3\): \(f(3) = \frac{9-3}{3-2} = 6\).
Since \(f \to \pm\infty\) at \(x = 2\) and as \(x \to \pm\infty\), there are no absolute extrema.

Key Concepts

Increasing and Decreasing Intervals

Understanding whether a function is increasing or decreasing over certain intervals involves looking at the function's derivative. A function is increasing on an interval if its derivative is positive throughout that interval. Conversely, a function decreases where its derivative is negative.

After calculating the derivative of the given function, it's important to find where this expression equals zero or does not exist, as these points can signify changes in direction. These are known as critical points. Once these critical points are identified, evaluate the sign of the derivative in each subinterval. If the derivative changes from positive to negative, the function switches from increasing to decreasing, and vice versa. This method allows us to map out where the function increases and decreases over its domain.

• Find the critical points: Set the derivative equal to zero and solve for the variable.
• Determine the intervals: Analyze the sign of the derivative in each interval divided by the critical points.
• Identify increasing/decreasing behavior: Positive derivative means the function is rising, negative means it's falling.

Local Extreme Values

In calculus, local extreme values refer to the highest or lowest points in a specific region of a function. These can be either local maxima (highest point) or local minima (lowest point). Finding these extremes relies on the function's derivative.

Critical points not only help determine increasing and decreasing intervals but also local extremes. Specifically, if a function's derivative changes sign from positive to negative at a critical point, that indicates a local maximum. If it changes from negative to positive, there's a local minimum. Calculating the derivative helps determine where these changes occur. Together with evaluating the original function, you can identify exactly where on the graph these points exist.

• Find and evaluate critical points: Check where the derivative changes sign.
• Determine maxima and minima: A positive to negative change indicates a local max; negative to positive indicates a local min.
• Analyze the function: Plug back into the original function to confirm the exact value.

Absolute Extreme Values

Absolute extreme values are the highest and lowest points over the entire domain of a function, not just within a particular region. Unlike local extremes, absolute extremes may occur at endpoints or within the domain where the derivative is zero or undefined.

To find these, calculate the value of the function at all critical points and endpoints. Compare these values to determine which are the global maximum and minimum. The function's absolute maximum is the largest function value obtained, while the absolute minimum is the smallest.

• Identify all critical points: Look for where derivatives are zero or undefined.
• Evaluate the function: Calculate the function's value at these points.
• Compare values: Determine which are highest and lowest across the entire range.

Being able to discern these points from a graph can also be crucial, especially if endpoints are involved in discrete domains.

Derivative Calculations

Derivative calculations are a cornerstone of solving calculus problems, including understanding function behaviors such as increasing and decreasing intervals and finding extreme values. For the given function, the derivative was calculated using the quotient rule. This rule applies when dividing two functions and is expressed as the derivative of the numerator times the denominator minus the derivative of the denominator times the numerator, all divided by the denominator squared.

The use of the quotient rule in solving the given problem demonstrates its practical application in calculus to understand vital characteristics of a function. Mastering derivatives opens up a plethora of capabilities in mathematical modeling and problem-solving.

• Quotient Rule: Apply when differentiating a fraction of functions.
• Chain Rule and Product Rule: Other common methods for derivative calculations based on function type.
• Application: Enables determining critical points, extrema, and intervals of behavior effectively.

Familiarity with different derivative rules enhances one's ability to tackle a wide range of calculus problems confidently.