Problem 42
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. $$f(x)=7$$
Step-by-Step Solution
Verified Answer
The difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x)=7\) simplifies to 0.
1Step 1: Substitute
Substitute \(f(x) = 7\) and \(f(x+h)\) into the quotient: \(\frac{f(x+h)-f(x)}{h} = \frac{7-7}{h}\).
2Step 2: Simplify
Simplify the numerator: \(\frac{7-7}{h} = \frac{0}{h}\).
3Step 3: Simplify
When zero is divided by any non-zero number the result is zero, which means that \(\frac{0}{h}=0\).
Key Concepts
Algebraic SimplificationRational ExpressionsConstant Functions
Algebraic Simplification
Algebraic simplification is a process used to make complex expressions easier to work with by reducing them to their simplest form. This process often involves combining like terms, reducing fractions, and applying the distributive property, among other techniques.
For the exercise of finding the difference quotient \( \frac{f(x+h)-f(x)}{h} \), algebraic simplification comes into play when we reduce the expression to its simplest form. As seen in the provided solution, after substituting the values of \( f(x) \) and \( f(x+h) \) into the difference quotient and performing the subtraction \( 7 - 7 \), we are left with a numerator of 0. Once we have \( \frac{0}{h} \), algebraic simplification dictates that any number divided by another number (except for 0) results in 0. This is an example of how simplification can lead to a more understandable result in algebra.
For the exercise of finding the difference quotient \( \frac{f(x+h)-f(x)}{h} \), algebraic simplification comes into play when we reduce the expression to its simplest form. As seen in the provided solution, after substituting the values of \( f(x) \) and \( f(x+h) \) into the difference quotient and performing the subtraction \( 7 - 7 \), we are left with a numerator of 0. Once we have \( \frac{0}{h} \), algebraic simplification dictates that any number divided by another number (except for 0) results in 0. This is an example of how simplification can lead to a more understandable result in algebra.
Rational Expressions
Rational expressions are fractions that involve polynomials in their numerator, denominator, or both. In algebra, they are akin to ratios of polynomials and require particular attention when it comes to simplification and operations such as addition, subtraction, multiplication, and division.
The difference quotient in algebra is often a rational expression, as it involves a numerator and a denominator. In our exercise, the rational expression \( \frac{7-7}{h} \) simplifies to \( \frac{0}{h} \). When dealing with rational expressions, it's important to identify opportunities to simplify, just like reducing \( \frac{0}{h} \) to 0 because a zero numerator indicates the whole expression is 0, as long as the denominator is not also zero. Simplifying rational expressions is essential for working with complex algebraic equations effectively.
The difference quotient in algebra is often a rational expression, as it involves a numerator and a denominator. In our exercise, the rational expression \( \frac{7-7}{h} \) simplifies to \( \frac{0}{h} \). When dealing with rational expressions, it's important to identify opportunities to simplify, just like reducing \( \frac{0}{h} \) to 0 because a zero numerator indicates the whole expression is 0, as long as the denominator is not also zero. Simplifying rational expressions is essential for working with complex algebraic equations effectively.
Constant Functions
Constant functions are a type of mathematical function where the value of the function is the same for any input. In other words, no matter what \( x \) value you choose, the output will always be the same, constant value. These functions are graphically represented as a horizontal line on the Cartesian plane.
In the context of the exercise, the function \( f(x) = 7 \) is a constant function because for any input \( x \), the output is always 7. This is why the difference quotient \( \frac{f(x+h)-f(x)}{h} \) simplifies to 0. The value of the function does not change with different \( x \) values, so \( f(x+h) \) and \( f(x) \) are equal, leading to a difference of zero. Understanding constant functions is helpful when determining the behavior of a function and predicting its graphical representation.
In the context of the exercise, the function \( f(x) = 7 \) is a constant function because for any input \( x \), the output is always 7. This is why the difference quotient \( \frac{f(x+h)-f(x)}{h} \) simplifies to 0. The value of the function does not change with different \( x \) values, so \( f(x+h) \) and \( f(x) \) are equal, leading to a difference of zero. Understanding constant functions is helpful when determining the behavior of a function and predicting its graphical representation.
Other exercises in this chapter
Problem 42
Give the slope and y-intercept of each line whose equation is given. Then graph the line. $$y=-3 x+2$$
View solution Problem 42
Express the given function h as a composition of two functions f and g so that \(h(x)=(f \circ g)(x)\) $$h(x)=\sqrt{5 x^{2}+3}$$
View solution Problem 42
Give the center and radius of the circle described by the equation and graph each equation. $$ x^{2}+y^{2}=49 $$
View solution Problem 43
Begin by graphing the standard cubic function, \(f(x)=x^{3} .\) Then use transformations of this graph to graph the given function. $$ r(x)=(x-3)^{3}+2 $$
View solution