Problem 110 - Precalculus : Building Concepts and Connections

Question

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=\frac{1}{x+1}, x \neq-1$$

Step-by-Step Solution

Answer

The difference quotient of the function \(f(x)=\frac{1}{x+1}\) is \(-\frac{1}{(x+1)(x+h+1)}\)

Step 1: Substituting the values

Substitute \(x+h\) into \(f(x)=\frac{1}{x+1}\) which makes it \(f(x+h) = \frac{1}{x + h + 1}\)

Step 2: Calculate the difference

The difference \(f(x+h) - f(x)\) will be \(\frac{1}{x + h + 1} - \frac{1}{x+1}\)

Step 3: Form the difference quotient

Form the difference quotient \(\frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x + h + 1} - \frac{1}{x+1}}{h}\)

Step 4: Simplify the expression

Simplify \(\frac{\frac{1}{x + h + 1} - \frac{1}{x+1}}{h}\) to \(\frac{(x+1) - (x + h + 1)}{h((x+1)(x+h+1))} = \frac{-h}{h((x+1)(x+h+1))}\)

Step 5: Cancel out the common terms

Cancel \(h\) from numerator and denominator of \(\frac{-h}{h((x+1)(x+h+1))}\) to get the expression \(-\frac{1}{(x+1)(x+h+1)}\)

Key Concepts

Difference Quotient Calculations

The difference quotient is a fundamental concept in calculus that measures how a function's value changes as its input changes. The formula for the difference quotient is \(\frac{f(x+h)-f(x)}{h}\), where \(h\) is the amount of change in the input value, and \(f(x)\) is the value of the function at \(x\). To calculate the difference quotient for a given function, you must perform the following steps: first, substitute \(x+h\) into the function to obtain \(f(x+h)\). Then, compute the difference \(f(x+h) - f(x)\), and finally, divide this difference by \(h\), with the condition that \(h\) is not equal to zero, as dividing by zero is undefined in mathematics.
Understanding the difference quotient is essential because it lays the groundwork for the derivative, a core concept in calculus which measures the instantaneous rate of change of a function. Essentially, as \(h\) approaches zero, the difference quotient gives the slope of the tangent line to the curve of the function at the point \(x\), which is the derivative of the function.

Algebraic Simplification

Algebraic simplification is the process of reducing expressions to their simplest form, making them easier to interpret or further manipulate. When working with the difference quotient, simplification often involves combining like terms, factoring, and canceling out common factors. In our exercise, after forming the difference quotient \(\frac{f(x+h) - f(x)}{h}\), we reach a complex fraction which needs simplification. The key to simplifying the expression is to find a common denominator for the two fractions within the numerator.
In our case, once you substitute \(x+h\) and \(x\) into \(f\) and calculate the difference, you'll have a fraction where both the numerator and the denominator contain an \(h\) term. Simplification then involves factoring out this common \(h\) and canceling it, a step often overlooked by students unfamiliar with algebraic identities and properties. Such cancellation is permitted because the terms are multiplied, not added or subtracted; this is why simplification of this nature is crucial for correct and clear results.

Rational Functions

A rational function is defined as the ratio of two polynomials. It is in the form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not the zero polynomial. The function \(f(x) = \frac{1}{x+1}\) from the exercise is an example of a rational function where the numerator is the constant 1, a polynomial of degree 0, and the denominator is \((x+1)\), a polynomial of degree 1.
The characteristics of rational functions, such as asymptotic behavior, intercepts, and discontinuities, are important to understand when analyzing and simplifying them. In the context of the difference quotient, it's important to highlight that the function cannot be evaluated at values that make the denominator zero, as indicated by the condition \(x eq -1\) in the given exercise. In the solution process for the difference quotient, when we approach the final expression \(-\frac{1}{(x+1)(x+h+1)}\), it represents how the original rational function changes around any point except \(x = -1\), maintaining the function's properties without causing division by zero.