Problem 43 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=\frac{x+3}{x-2} $$

Step-by-Step Solution

Answer

The function has no relative extrema.

Step 1: Find the Derivative

To find the relative extrema, we begin by finding the first derivative of the function. Given \( f(x) = \frac{x+3}{x-2} \), we use the quotient rule for differentiation, \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x+3 \) and \( v = x-2 \), so \( u' = 1 \) and \( v' = 1 \). Calculating the derivative: \[ f'(x) = \frac{(x-2)(1) - (x+3)(1)}{(x-2)^2} = \frac{x-2-x-3}{(x-2)^2} = \frac{-5}{(x-2)^2} \]

Step 2: Set the Derivative to Zero

To find critical points, set the first derivative equal to zero and solve for \( x \). The derivative is \( f'(x) = \frac{-5}{(x-2)^2} \). Since the numerator \(-5\) is a constant, there are no values of \( x \) for which \( f'(x) = 0 \).

Step 3: Identify Undefined Points of the Derivative

The critical points can also occur where the derivative is undefined. The expression \( (x-2)^2 = 0 \) will cause \( f'(x) \) to be undefined when \( x = 2 \). At \( x = 2 \), the original function is also undefined, so \( x = 2 \) is a vertical asymptote, not a point of relative extremum.

Step 4: Determine Relative Extrema

Since there are no solutions where the derivative equals zero and the only undefined point corresponds to a vertical asymptote, the function \( f(x) = \frac{x+3}{x-2} \) has no relative extrema.

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on change. It helps us understand how functions behave.
In the context of finding relative extrema, we often investigate the behavior of a function's derivative.
The derivative can reveal information about where a function increases or decreases. It can also tell us about peaks and valleys. These are known as relative maxima and minima, or extremum points.
For a function, relative extrema occur where the slope of the tangent (the derivative) is zero or undefined. These are known as critical points. From such points, one can determine where the function reaches its highest or lowest local value. Thus, calculus provides tools to analyze and predict changes in varied phenomena. It's fundamental for both theoretical and practical applications.

Differentiation

Differentiation is a process in calculus. It involves finding a derivative of a function.
The derivative represents the rate at which a function is changing at any given point. It functions as a key player in determining slopes, rates, and motion.

• How to Differentiate: To differentiate, we often use rules like the product rule, chain rule, and quotient rule.
• Critical Points Analysis: Once you have the derivative, solving for where it equals zero helps find critical points.
• Further Analysis: Assess where the derivative is undefined or doesn't exist; these points may reflect behavior such as asymptotic tendencies.

In our exercise, differentiation led us to find that the derivative of the function is never zero. Thus, we identified that the only place where the derivative is undefined is crucially due to a vertical asymptote. Differentiation reveals no relative extrema in this scenario, which is essential for characterizing the nature of this function's graph.

Quotient Rule

The quotient rule is essential when differentiating functions expressed as a ratio.
This concept comes into play whenever we have a function divided by another function. The rule is mathematically structured by this formula:
If we have \( f(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are functions of \( x \), then:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]
This formula is instrumental for checking how a change in \( x \) influences the entire ratio. Understanding this is vital for problems involving ratios and fractions.

• Using the Quotient Rule: Identify the numerator and the denominator as separate functions \( u \) and \( v \) respectively, compute their derivatives \( u' \) and \( v' \).
• Plug into the Formula: Substitute these into the quotient rule to find the overall derivative.
• Application in Exercise: In our specific problem, applying the quotient rule revealed the derivative \( f'(x) = \frac{-5}{(x-2)^2} \).

Ultimately, the quotient rule helps us discover how each small change in one part of the expression influences the function's rate of change overall. This vital concept allows us to tackle problems involving real-world situations represented by fraction-based functions.