Problem 13 - Single Variable Calculus

Question

Use the definition of the derivative to find the derivative of the function. What is its domain? \(f(x)=\frac{3}{2 x+1}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\frac{3}{2x + 1}\) is \(f'(x) = \frac{-6}{(2x + 1)^2}\) and the domain is all real numbers except \(x = -\frac{1}{2}\).

Step 1: Find f(x+h)

First, we will find \(f(x+h)\), where \(f(x) = \frac{3}{2x +1}\). \(f(x+h) = \frac{3}{2(x + h) + 1}\)

Step 2: Calculate f(x+h) - f(x)

We need to find the difference between \(f(x+h)\) and \(f(x)\). \(\Delta f = f(x+h) - f(x) = \frac{3}{2(x + h) + 1} - \frac{3}{2x + 1}\)

Step 3: Calculate the limit

Using the limit definition of the derivative, we will find the derivative of \(f(x)\). \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\Delta f}{h}\) To calculate this limit, we need to first simplify \(\Delta f\) and after that divide it by \(h\).

Step 4: Simplify the expression

First, we find a common denominator and simplify \(\Delta f\). \(\Delta f = \frac{3(2x + 1) - 3(2x + 2h + 1)}{(2x + 1)(2(x + h) + 1)}\) \(\Delta f = \frac{-6h}{(2x+1)(2(x + h) + 1)}\) Now, we divide this expression by \(h\): \(\frac{\Delta f}{h} = \frac{-6}{(2x + 1)(2(x + h) + 1)}\)

Step 5: Calculate the limit using the simplified expression

Now, we can find the limit: \(f'(x) = \lim_{h \to 0} \frac{-6}{(2x + 1)(2(x + h) + 1)} = \frac{-6}{(2x + 1)^2}\)

Step 6: Determine the domain of the derivative

Finally, we need to find the domain of the derivative. The function is undefined when the denominator of the derivative is equal to 0. Therefore, we need to solve for x: \((2x + 1)^2 = 0\) \(2x + 1 = 0\) \(x = -\frac{1}{2}\) Since the original function and its derivative are undefined when \(x = -\frac{1}{2}\), we exclude this value from the domain. Thus, the domain of the derivative is: \(-\infty \lt x \lt -\frac{1}{2} \) or \(-\frac{1}{2} \lt x \lt \infty\) In summary, the derivative of the function \(f(x)=\frac{3}{2x + 1}\) is \(f'(x) = \frac{-6}{(2x + 1)^2}\) and the domain is all real numbers except \(x = -\frac{1}{2}\).

Key Concepts

Limit Definition

The concept of limits is a fundamental building block in calculus that helps us understand the behavior of functions as they approach specific points. The limit definition of a derivative is critical to finding the rate of change or the slope of the function at a particular point.
Using the limit definition means calculating:

• The derivative of a function, expressed as\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]

This formula examines the value of the function very near the point of interest. As \(h\), a small increment, approaches zero, the difference quotient \(\frac{f(x+h) - f(x)}{h}\) approximates the derivative.
By simplifying this expression and calculating its limit, we find the derivative which provides deep insights into the behavior of the function around that point. This process can sometimes be complex, requiring algebraic manipulation to handle fractions properly, just like in our original problem.

Domain of a Function

The domain of a function includes every allowable input value for which the function is defined. For rational functions, the domain is all real numbers except where the denominator equals zero because division by zero is undefined.

• To find the domain, identify where the function cannot be evaluated.
• For \(f(x) = \frac{3}{2x + 1}\), the denominator \(2x + 1\) must not be zero.

Solve \(2x + 1 = 0\) to find the restricted values: \(x = -\frac{1}{2}\). Thus, the domain of the function is:

• All real numbers except \(x = -\frac{1}{2}\)

Rational functions have continuous behavior everywhere in their domain, giving them distinct characteristics and making them interesting to analyze.

Rational Functions

Rational functions are quotients of two polynomial functions. In our case, the function \(f(x) = \frac{3}{2x+1}\) is a simple rational function.
These functions have characteristics primarily governed by:

• The expressions in the numerator and denominator.
• Vertical asymptotes, which occur where the denominator is zero.

Such functions can have breaks or holes at these asymptotes, reflecting where the function is undefined. Understanding rational functions involves:

• Identifying asymptotes.
• Analyzing end behavior to see how they approach infinity.

Despite their complexity, rational functions can be simplified using algebraic techniques, allowing us to find derivatives and domain constraints. In the original problem, using simplification through the limit definition of a derivative, we attain the derivative: \(f'(x) = \frac{-6}{(2x + 1)^2}\), demonstrating deeper analysis of the rational function's behavior near its points of interest.