Chapter 3: Problem 26
(a) Find the difference quotient \(\frac{f(x)-f(a)}{x-a}\) for each function, as in Example 4. (b) Find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for each function, as in Example \(5 .\) $$f(x)=-2 x+5$$
Short Answer
Expert verified
The difference quotient in both cases is \(-2\).
Step by step solution
01
Difference Quotient Setup for (a)
We are given the function \( f(x) = -2x + 5 \). For part (a), to find the difference quotient \( \frac{f(x) - f(a)}{x - a} \), start by calculating \( f(a) \). This means substituting \( a \) in place of \( x \) in the function, which gives: \( f(a) = -2a + 5 \).
02
Simplify the Difference Quotient for (a)
Substitute \( f(a) = -2a + 5 \) and \( f(x) = -2x + 5 \) back into the expression for the difference quotient: \[ \frac{f(x) - f(a)}{x - a} = \frac{(-2x + 5) - (-2a + 5)}{x - a} \] \[ = \frac{-2x + 5 + 2a - 5}{x - a} \] \[ = \frac{-2x + 2a}{x - a} \] \[ = \frac{2(a - x)}{x - a} \] \[ = -2 \] Here, the terms \( 5 \) cancel out, and dividing \( 2(a-x) \) by \( x-a \) results in \(-2\).
03
Difference Quotient Setup for (b)
Now, for part (b), we need to find the difference quotient \( \frac{f(x+h) - f(x)}{h} \). First, calculate \( f(x+h) \) by substituting \( x+h \) into the function: \( f(x+h) = -2(x + h) + 5 = -2x - 2h + 5 \).
04
Simplify the Difference Quotient for (b)
Substitute \( f(x) = -2x + 5 \) and \( f(x+h) = -2x - 2h + 5 \) into the expression for the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{(-2x - 2h + 5) - (-2x + 5)}{h} \] \[ = \frac{-2x - 2h + 5 + 2x - 5}{h} \] \[ = \frac{-2h}{h} \] \[ = -2 \] Here, the \( 2x \) and \( 5 \) terms cancel out, and dividing \( -2h \) by \( h \) simplifies to \(-2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Functions
A function in mathematics is like a special machine that takes an input, usually denoted by a variable like \( x \), and outputs a single result. This relation is defined by an expression or rule, such as \( f(x) = -2x + 5 \) in our example.
Functions are widely used in algebra and calculus to express mathematical relationships. In the equation \( f(x) = -2x + 5 \), the function \( f \) assigns to each value of \( x \) an output value.
Key points about functions include:
Functions are widely used in algebra and calculus to express mathematical relationships. In the equation \( f(x) = -2x + 5 \), the function \( f \) assigns to each value of \( x \) an output value.
Key points about functions include:
- The input \( x \) can be any value that the function is designed to handle, also known as the domain of the function.
- The output is the range, what you get after substituting \( x \) into the function equation.
Algebra
Algebra is the branch of mathematics that deals with symbols and the rules for manipulating those symbols. It is a unifying thread of almost all of mathematics and it is about finding the unknown or putting real-life variables into equations and then solving them.
The difference quotient \( \frac{f(x)-f(a)}{x-a} \) is a part of algebra that deals with functions. It represents the average rate of change of the function \( f \) between two points \( x \) and \( a \). The process involves:
The difference quotient \( \frac{f(x)-f(a)}{x-a} \) is a part of algebra that deals with functions. It represents the average rate of change of the function \( f \) between two points \( x \) and \( a \). The process involves:
- Identifying the function, which in our case is \( f(x) = -2x + 5 \).
- Calculating the change in the function values \( f(x) - f(a) \).
- Dividing by the change in \( x \), i.e., \( x - a \) to get the average rate of change.
Precalculus
Precalculus is a course with mathematical techniques necessary for the study of calculus. It covers important concepts, building a solid foundation to prepare for understanding limits, derivatives, and integrals.
One crucial concept in precalculus is the difference quotient, which introduces the idea of derivatives. It's a precursor to the derivative since it gives the average rate of change of a function over an interval.
For instance, in our example \( f(x) = -2x + 5 \), the difference quotient \( \frac{f(x+h)-f(x)}{h} \) calculates how much the function \( f(x) \) changes as \( x \) changes by a small amount \( h \).
One crucial concept in precalculus is the difference quotient, which introduces the idea of derivatives. It's a precursor to the derivative since it gives the average rate of change of a function over an interval.
For instance, in our example \( f(x) = -2x + 5 \), the difference quotient \( \frac{f(x+h)-f(x)}{h} \) calculates how much the function \( f(x) \) changes as \( x \) changes by a small amount \( h \).
- This quotient simplifies to \(-2\), indicating a constant rate of change, which is typical for linear functions.
- In general, as \( h \) approaches zero, this quotient leads to the derivative, which tells how the function \( f \) changes at each point directly.