Problem 41 - College Algebra

Question

In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=6$$

Step-by-Step Solution

Answer

In conclusion, the simplified form of the difference quotient for the function \(f(x)=6\) is 0.

Step 1: Substitute the function into the difference quotient

Substitute \(f(x) = 6\) into the difference quotient. This translates into: \[\frac{f(x+h)-f(x)}{h} = \frac{6-6}{h}.\]

Step 2: Simplify the numerator

The numerator simplifies into 0 because 6 - 6 = 0. This results in: \[\frac{0}{h}.\]

Step 3: Simplify the difference quotient

Any number divided by 0 is typically undefined. However, since 0 is in the numerator, \[\frac{0}{h} = 0.\]

Key Concepts

Function Substitution

Function substitution is a fundamental concept in algebra, particularly when working with difference quotients. When you have a function, say, \(f(x) = 6\), it means that no matter the input, the output will always be 6. To find the difference quotient for this constant function, you begin by substituting the function into the quotient formula. This is a straightforward step where you replace \(f(x + h)\) and \(f(x)\) with 6, since the function's value does not depend on \(x\) or any other variable. So, the expression becomes \(\frac{6-6}{h}\). This substitution is crucial because it translates the abstract difference quotient into something tangible we can work with.

Simplifying Expressions

Simplifying expressions is about reducing them to their most basic form. Once you've substituted the function into the difference quotient formula, the next step is simplification. For instance, in the expression \(\frac{6-6}{h}\), the numerator simplifies as \(6 - 6\). This operation is simple as it resolves to zero, leading to the simplified expression \(\frac{0}{h}\). By simplifying the expressions at each step, you eliminate unnecessary complexity and make solving the problem much more straightforward. It's like cleaning up your workspace to easily see what you’re dealing with.

Constant Function

A constant function is a type of function where the output value is the same regardless of the input. In mathematical terms, \(f(x) = c\), where \(c\) is a constant. For example, \(f(x) = 6\) is a constant function because no matter what value \(x\) takes, \(f(x)\) will always be 6. This property greatly affects calculations such as the difference quotient since the change in the function, \(f(x+h) - f(x)\), will always equal zero. Recognizing a constant function quickly allows you to predict that simplifications involving the difference quotient will often result in zero, as seen when the numerator equals zero. Understanding constant functions boosts efficiency when solving similar algebraic problems.

Numerator and Denominator

In any fraction, there are two critical parts: the numerator and the denominator. Understanding their roles is key in mathematics. In the difference quotient \(\frac{f(x+h)-f(x)}{h}\), the numerator is \(f(x+h)-f(x)\) and the denominator is \(h\). Here, the numerator expresses the difference in the function values, while the denominator represents the change in \(x\). For constant functions like \(f(x) = 6\), the difference \(f(x+h)-f(x)\) becomes zero, simplifying our work to \(\frac{0}{h}\). The "zero" numerator dominates the simplification process, making the outcome equivalent to zero, as zero divided by any non-zero number (h in this case) is still zero. Understanding how numerators and denominators interact helps in solving and simplifying algebraic fractions seamlessly.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to solve problems or make them easier to understand. Once you've substituted and simplified expressions like \(\frac{6-6}{h}\) to \(\frac{0}{h}\), you use algebraic principles to determine that this equals zero. Because any number multiplied or divided by zero results in zero, we bypass the complexities often involved in undefined terms because the numerator is zero. This manipulation highlights the power of algebraic rules, enabling the transformation of daunting expressions into simpler, logically sound forms. Thus, understanding algebraic manipulation aids in tackling more intricate equations as it provides a toolkit for simplifying challenges systematically.