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In Exercises \(45-52,\) find the average rate of change of the function from \(x=1\) to \(x=3\) $$f(x)=x^{3}$$

Short Answer

Expert verified
The average rate of change is 13.

Step by step solution

01

Identify the Formula for Average Rate of Change

The average rate of change of a function 1 between two points, 2 and 3, is given by the formula: \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \). Here, \( x_1 = 1 \) and \( x_2 = 3 \).
02

Calculate the Function Values

First, find \( f(1) \) and \( f(3) \). Since \( f(x) = x^3 \), we have: \( f(1) = 1^3 = 1 \) and \( f(3) = 3^3 = 27 \).
03

Substitute Values into the Formula

Substitute \( f(1) = 1 \), \( f(3) = 27 \), \( x_1 = 1 \), and \( x_2 = 3 \) into the average rate of change formula: \( \frac{27 - 1}{3 - 1} \).
04

Simplify the Expression

Simplify the expression obtained in the previous step: \( \frac{26}{2} = 13 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Functions
Functions are one of the most fundamental concepts in mathematics and are essential for describing relationships between varying quantities. The idea of a function is to link each input value to a unique output value. In simple terms, think of a function as a machine where you input a number, and it gives you another number as output.

For any function, like the one in our exercise, which is defined as
  • \(f(x) = x^3\)
you insert an \(x\) value (like 1 or 3) and out comes another number, which in our case is the cube of \(x\). So, plug in 1, and the function returns 1. Plug in 3, and you get 27.

This concept is powerful because it helps us predict and understand how things change. In real life, functions can model everything from the population growth in a city to the cooling of a hot cup of coffee. Mastering functions is key to advancing in math and science, as they form the backbone of many complex concepts.
Polynomials
Polynomials are an exciting part of algebra and mathematics. They are expressions of sums of powers in one or more variables multiplied by coefficients. A polynomial can be as simple as a single number or a complex expression like
  • \(x^3\)

In our example, we are dealing with a cubic polynomial, given by:
  • \(f(x) = x^3\)
Polynomials are fun because they have unique properties, like their degree, which tells you the highest power of the variable. Here, the degree is 3, meaning the graph of the function will have a certain curve drawing an interesting shape on a plot.

Working with polynomials also involves various operations, such as addition, subtraction, multiplication, and even division. However, the most common application is finding roots, critical points, and, as in our exercise, discussing how they change across intervals by using tools like the average rate of change. Understanding polynomials opens doors to deeper mathematical study.
Calculus Concepts
Calculus introduces a whole new dimension to how we understand change and motion. One of the first ideas you encounter in calculus is the average rate of change, as seen in this exercise. Simply put, it's a way to measure how fast something is changing over a certain interval.

Using calculus terminology, the formula applied here was
  • \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \)
This formula perfectly captures the essence of change. When you calculate the difference in function values (or outputs) and divide it by the change in \(x\) values (inputs), you get an average rate. It's like finding the speed of a journey by dividing the total distance by the total time.

Calculus is the study of change, and it helps you understand deeply how functions behave across intervals. Knowing how to determine the average rate of change is your stepping stone toward more complex calculus concepts like derivatives, which further describe how rates change instantaneously. By familiarizing yourself with these foundational ideas, you prepare yourself for a world of mathematical exploration.

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Most popular questions from this chapter

Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price \(p\) and the number of units for sale \(x\) are related by the demand equation: $$p=3000-\frac{1}{2} x$$ Assume that the cost \(C(x)\) of producing \(x\) items is governed by the equation $$C(x)=2000+10 x$$ and the revenue \(R(x)\) generated by selling \(x\) units is governed by $$R(x)=100 x$$ a. Write the cost as a function of price \(p\) b. Write the revenue as a function of price \(p\) c. Write the profit as a function of price \(p\)

For Exercises \(111-114\), refer to the following: In calculus, the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of a function \(f\) is used to find the derivative \(f^{\prime}\) of \(f\), by allowing \(h\) to approach zero, \(h \rightarrow 0 .\) Find the derivative of the following functions. \(f(x)=a x^{2}+b x+c,\) where \(a, b,\) and \(c\) are constants, \(a \neq 0\)

Given the function \(f(x)=x^{2}+7\) and \(g(x)=\sqrt{x-3},\) find \(f \circ g,\) and state the domain. $$\begin{aligned}\text { Solution: } f \circ g &=f(g(x))=(\sqrt{x-3})^{2}+7 \\\&=f(g(x))=x-3+7 \\\&=x-4\end{aligned}$$ Domain: \((-\infty, \infty)\) This is incorrect. What mistake was made?

Plot the function \(f(x)=\frac{\sin x}{\cos x} .\) What function is this?

Graph the functions \(f\) and \(g\) and the line \(y=x\) in the same screen. Do the two functions appear to be inverses of each other? $$f(x)=\sqrt{4-3 x} ; \quad g(x)=\frac{4}{3}-\frac{x^{2}}{3}, x \geq 0$$

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