Problem 23 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 21 through 24 use the limit definition of \(f^{\prime}(a)\) to find the derivative of \(f\) at the point indicated. $$ f(x)=\frac{x}{2}+\frac{2}{x} \text { at } x=1 $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\frac{x}{2}+\frac{2}{x}\) at the point \(x=1\) is \(-\frac{3}{2}\).

Step 1: Apply the Limit Definition of the Derivative

First, apply the definition of the derivative, which is \(f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\). Substituting the function \(f(x)=\frac{x}{2}+\frac{2}{x}\) and the point \(x=1\) into the definition, we get \[f'(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{\left(\frac{1+h}{2}+\frac{2}{1+h}\right)-\left(\frac{1}{2}+\frac{2}{1}\right)}{h}\]

Step 2: Simplify the Fraction Inside the Limit

Next, simplify the expression in the numerator. Do this by combining like terms and simplifying the fractions:\[\lim _{h \rightarrow 0} \frac{\left(\frac{1+h}{2}+\frac{2}{1+h}-\frac{1}{2}-2\right)}{h}=\lim _{h \rightarrow 0} \frac{\left(\frac{h}{2}-\frac{2h}{1+h}\right)}{h}\] Then, divide both terms in the numerator by \(h\):\[\lim _{h \rightarrow 0} \left(\frac{1}{2}-\frac{2}{1+h}\right)\]

Step 3: Evaluate the Limit

Finally, find the limit as \(h\) approaches zero: \[f'(1)=\frac{1}{2}-\frac{2}{1+0}=\frac{1}{2}-2=-\frac{3}{2}\]

Key Concepts

Finding Derivatives

The very idea of finding derivatives in calculus revolves around understanding how a function changes at any given point. We can think of derivatives as the 'instantaneous rate of change' of a function. This rate of change can be found using the limit definition of a derivative.

In our example, we used the function \( f(x)=\frac{x}{2}+\frac{2}{x} \) at \( x=1 \). To find the derivative \( f'(x) \), we applied the definition formula:

• Step 1 involved substituting the required function values into the limit formula: \( f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} \).
• Step 2 required manipulating and simplifying the expression to get to the final form that we could evaluate.
• Finally, Step 3 was taking the limit by substituting \( h = 0 \).

Each step simplifies the process, making derivatives easier to understand and calculate.

Calculus Problems

Solving calculus problems often involves multiple steps, and derivatives are no exemption. These problems allow us to determine how a slight change in input affects the output of a function. Sectioning problems into smaller, digestible steps can aid in understanding.

For example, the function \( f(x)=\frac{x}{2}+\frac{2}{x} \) demonstrates common calculus practices, such as breaking down complex fractions and using limits to simplify expressions.

Breaking down calculus problems using:

• The application of mathematical rules, such as limits or function properties, to obtain the derivatives.
• Utilizing algebraic shortcuts to make the simplification process less tedious.
• Evaluating the results accurately to ensure each step is correctly followed.

These approaches help maintain clarity and understandability.

Functions and Their Rates of Change

Functions are math’s way of representing real-world activities, and understanding how these functions change is crucial. A derivative tells us how fast a function is changing at any given point, which is crucial in fields like physics, engineering, and economics.

In the given problem, \( f(x)=\frac{x}{2}+\frac{2}{x} \) at \( x=1 \), the derivative \( f' (1)\) represents the rate at which the function value changes at that specific point. It describes:

• The slope of the tangent line to the function's graph at \( x=1 \).
• The direction of change, with the sign of the derivative clarifying if the function is increasing or decreasing.
• Whether the point signifies a maximum, minimum, or point of inflection.

By understanding this, students can appreciate the role derivatives play in analytical processes and real-life applications.