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(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:
  • 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.
Understanding how a function works provides a foundation for more complex concepts like calculus and helps in solving problems that involve changing quantities, which are common in the real world.
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:
  • 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.
Algebra is essential not just for understanding functions, but it's also crucial for solving equations and understanding mathematical relationships related to rate and speed.
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 \( h \).
  • 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.
The techniques learned in precalculus are critical for solving real-world problems where it's important to understand how a system is changing, not just its static state.

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Most popular questions from this chapter

The \(3 x+1\) conjecture Define a function \(f,\) with domain the positive integers, as follows: $$f(x)=\left\\{\begin{array}{ll} 3 x+1 & \text { if } x \text { is odd } \\ x / 2 & \text { if } x \text { is even } \end{array}\right.$$ (a) Compute \(f(1), f(2), f(3), f(4), f(5),\) and \(f(6)\) (b) Compute the first three iterates of \(x_{0}=1\) (c) Compute the iterates of \(x_{0}=3\) until you obtain the value \(1 .\) [After this, the iterates will recycle through the simple pattern obtained in part (b). \(]\) (d) The \(3 x+1\) conjecture asserts that for any positive integer \(x_{0}\), the iterates eventually return to the value \(1 .\) Verify that this conjecture is valid for each of the following values of \(x_{0}: 2,4,5,6,\) and 7 Remark: At present, the \(3 x+1\) conjecture is indeed a conjecture, not a theorem. No one yet has found a proof that the assertion is valid for every positive integer. Computer checks, however, have verified the conjecture on a case-by-case basis for very large values of \(x_{0}\) As of April 2000, the conjecture had been verified for all values of \(x_{0}\) up to approximately \(1.8 \times 10^{16}\).

Use this definition: A prime number is a positive whole number with no factors other than itself and \(1 .\) For example, \(2,13,\) and 37 are primes, but 24 and 39 are not. \(B y\) convention 1 is not considered prime, so the list of the first few primes is as follows: \(2,3,5,7,11,13,17,19,23,29, \ldots\) Let \(f\) be the function that assigns to each natural number \(x\) the number of primes that are less than or equal to \(x .\) For example, \(f(12)=5\) because, as you can easily check, five primes are less than or equal to \(12 .\) Similarly, \(f(3)=2\) because two primes are less than or equal to \(3 .\) Find \(f(8)\) \(f(10),\) and \(f(50)\)

If \(g(x)=x^{2}-3 x k-4\) and \(g(1)=-2,\) find \(k\)

(a) Complete the given table. $$\begin{array}{l}x \quad x^{2} \quad(x-1)^{2} \quad(x+1)^{2} \\\\\hline 0 \\\1 \\\2 \\\3 \\\\-1 \\\\-2 \\\\-3 \\\\\hline\end{array}$$ (b) Using the results in the table, graph the functions \(y=x^{2}, y=(x-1)^{2},\) and \(y=(x+1)^{2}\) on the same set of axes. How are the graphs related?

$$\text { Let } k(x)=5 x^{3}+\frac{5}{x^{3}}-x-\frac{1}{x} . \text { Show that } k(x)=k(1 / x)$$

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