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Chapter 2: Problem 1

Find the average rate of change of the function over the given interval or intervals. \(f(x)=x^{3}+1\) a. [2,3] b. [-1,1]

Short Answer

Expert verified
Interval [2,3]: 19; Interval [-1,1]: 1.

Step by step solution

01

Understand the Average Rate of Change

The average rate of change of a function between two points is similar to finding the slope of the secant line that passes through the points y_1 = f(x_1) y_2 = f(x_2). It is calculated using the formula: \[\text{Average Rate of Change} \ = \ \frac{f(x_2) - f(x_1)}{x_2 - x_1}\]
02

Calculate Function Values for Interval [2,3]

The first interval is [2,3]. We need to calculate the values of the function at these points:\[f(x) = x^3 + 1\]Calculate \(f(2)\):\[f(2) = 2^3 + 1 = 8 + 1 = 9\]Calculate \(f(3)\):\[f(3) = 3^3 + 1 = 27 + 1 = 28\]
03

Calculate Average Rate of Change for Interval [2,3]

Now that we have \(f(2) = 9\) and \(f(3) = 28\), use the average rate of change formula:\[\frac{f(3) - f(2)}{3 - 2} = \frac{28 - 9}{1} = 19\]
04

Calculate Function Values for Interval [-1,1]

The second interval is [-1,1]. Calculate the function values at these points:Calculate \(f(-1)\):\[f(-1) = (-1)^3 + 1 = -1 + 1 = 0\]Calculate \(f(1)\):\[f(1) = 1^3 + 1 = 1 + 1 = 2\]
05

Calculate Average Rate of Change for Interval [-1,1]

Using \(f(-1) = 0\) and \(f(1) = 2\), apply the average rate of change formula:\[\frac{f(1) - f(-1)}{1 - (-1)} = \frac{2 - 0}{2} = 1\]
06

Final Results

The average rate of change for the interval [2,3] is 19, and for the interval [-1,1] is 1.

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Key Concepts

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

Secant Line
The concept of a secant line is fundamental in understanding the average rate of change. A secant line is a straight line that connects two points on the graph of a function. When you look at the function graph, the secant line's slope represents the rate at which the function changes between those two points.
For instance, if you draw a secant line over the interval r{}[2,3] for the function r{}\(f(x) = x^3 + 1\),
it touches the graph at the points (2, f(2)) and (3, f(3)). The slope of this line gives you the average rate of increase or decrease in the function across that interval. So, visualizing secant lines can make the concept of rate of change more intuitive.
Slope Formula
The slope formula is a handy tool for finding the steepness of a line, which in this context, helps us determine the average rate of change of a function. The formula is expressed as:\[\text{Slope} \ = \ \frac{y_2 - y_1}{x_2 - x_1}\]
In our exercise, the slope formula translates to evaluating r{}\(f(x_2) - f(x_1)\) and dividing it by \(x_2 - x_1\).
This assesses how much the function value (or y-value) changes per unit change in x.
  • For the interval [2,3], you substitute f(3) and f(2) into the formula to compute the slope as the average rate of change.
  • This simply applies the idea that each unit increase or decrease in the x-direction corresponds to a change in the function's output.
Understanding this helps you calculate and interpret changes accurately between any two points on a function curve.
Polynomial Functions
A polynomial function, like \(f(x) = x^3 + 1\), is comprised of terms that are powers of x, combined using addition, subtraction, and multiplication.
Polynomials can vary in complexity from linear equations to higher degree equations like cubic, quadratic, etc., depending on the highest power of x present. They are an essential part of algebra, as they are used to model a wide array of real-world scenarios.
  • Cubic polynomials, for example, have the form \(ax^3 + bx^2 + cx + d\), where the highest exponent is 3.
  • These functions will often have graphs that curve and change direction, which poses specific challenges for calculating rates of change, hence why the secant line is useful.
With polynomials, each term contributes uniquely to the overall behavior of the graph, influencing its valleys, peaks, and other critical points.
Interval Calculations
Performing interval calculations allows us to determine changes over specified portions of a function graph. It's vital to correctly identify these intervals, which is represented as \([a,b]\), where "a" and "b" denote the endpoints of the interval.
With the function \(f(x) = x^3 + 1\), evaluating it over different intervals illuminates how the function behaves between specific x-values.
  • For the interval \([2,3]\), we calculate the function values at x = 2 and x = 3. Then, using these values, we can apply the slope formula to find the average rate of change.
  • Similarly, for the interval \([-1,1]\), this process repeats to highlight how values change significantly over smaller or negative x-values.
Understanding interval calculations is crucial to applying the concept of average rate of change effectively across diverse problems.

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Most popular questions from this chapter

Graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$f(x)=\frac{\sin x}{|x|}$$

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ occur frequently in calculus. Evaluate this limit for the given value of \(x\) and function \(f.\) $$f(x)=x^{2}, \quad x=-2$$

Gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) $$0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon$$ $$f(x)=3-2 x, \quad c=3, \quad \epsilon=0.02$$

Use a CAS to perform the following steps: a. Plot the function near the point \(c\) being approached. b. From your plot guess the value of the limit. $$\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{x}$$

You will find a graphing calculator useful for Exercise. Let \(g(x)=\left(x^{2}-2\right) /(x-\sqrt{2}).\) a. Make a table of the values of \(g\) at the points \(x=1.4,1.41\) 1.414, and so on through successive decimal approximations of \(\sqrt{2} .\) Estimate \(\lim _{x \rightarrow \sqrt{2}} g(x).\) b. Support your conclusion in part (a) by graphing \(g\) near \(c=\sqrt{2}\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow \sqrt{2}.\) c. Find \(\lim _{x \rightarrow \sqrt{2}} g(x)\) algebraically.
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