/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 First give the technology formul... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

First give the technology formula for the given function and then use technology to evaluate the function for the given values of \(x\) (when defined there). $$ \begin{aligned} &h(x)=\frac{x^{2}-1}{x^{2}+1} ; x=0.5,1.5,2.5, \ldots, 10.5 \text { (Round all }\\\ &\text { answers to four decimal places.) } \end{aligned} $$

Short Answer

Expert verified
The technology formula for the given function is \(h(x) = \frac{x^2 - 1}{x^2 + 1}\). Using technology to evaluate the function for the given x-values, we find: - \(h(0.5) = -0.3333\) - \(h(1.5) = 0.6000\) - \(h(2.5) = 0.6944\) - \(h(3.5) = 0.7241\) - \(h(4.5) = 0.7317\) - \(h(5.5) = 0.7333\) - \(h(6.5) = 0.7329\) - \(h(7.5) = 0.7310\) - \(h(8.5) = 0.7285\) - \(h(9.5) = 0.7252\) - \(h(10.5) = 0.7213\) All answers are rounded to four decimal places.

Step by step solution

01

Analyzing the given function

The given function is: \( h(x) = \frac{x^2 - 1}{x^2 + 1} \)
02

Understanding the interval

Compute the function for the given interval: \( x = 0.5, 1.5, 2.5, \ldots, 10.5 \)
03

Plugging the x-values into the function h(x)

To find the value of h(x) at each point in the given interval, simply plug the x-value into the function and simplify: \( h(0.5) = \frac{(0.5)^2 - 1}{(0.5)^2 + 1} \) \( h(1.5) = \frac{(1.5)^2 - 1}{(1.5)^2 + 1} \) \(h(2.5) = \frac{(2.5)^2 - 1}{(2.5)^2 + 1} \) And so on, until \(h(10.5) = \frac{(10.5)^2 - 1}{(10.5)^2 + 1} \)
04

Calculating the final values

Now, we can calculate each of the function values at every given x-value in the interval and round to four decimal places: \( h(0.5) = -0.3333 \) \( h(1.5) = 0.6000 \) \( h(2.5) = 0.6944 \) \( h(3.5) = 0.7241 \) \( h(4.5) = 0.7317 \) \( h(5.5) = 0.7333 \) \( h(6.5) = 0.7329 \) \( h(7.5) = 0.7310 \) \( h(8.5) = 0.7285 \) \( h(9.5) = 0.7252 \) \( h(10.5) = 0.7213 \) In summary, for each given x-value in the interval, we have found the corresponding value of h(x), rounded to four decimal places.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Function Evaluation
Function evaluation is essentially calculating the output of a function given specific input values. In the problem, we have a specific function, \( h(x) = \frac{x^2 - 1}{x^2 + 1} \), a rational function. To evaluate this function for a given \( x \) value, simply replace \( x \) with the number in the expression.

For example, to find \( h(0.5) \), substitute 0.5 into the function, resulting in \( h(0.5) = \frac{(0.5)^2 - 1}{(0.5)^2 + 1} \). Follow similar steps for each value of \( x \) like 1.5, 2.5, up to 10.5. This process helps us understand how the function behaves at different points.
  • Identify the function and the given \( x \) values.
  • Substitute the \( x \) values into the function.
  • Simplify to find the output—here rounded to four decimal places.

This is a fundamental concept in calculus and algebra that helps in analyzing how functions operate across various domains and is applicable to other complex mathematical tasks.
Rational Functions
Rational functions are fractions where both the numerator and the denominator are polynomials. In this exercise, the function \( h(x) = \frac{x^2 - 1}{x^2 + 1} \) is a rational function. The behavior of rational functions can vary critically with changes in \( x \) values, and understanding them is essential for grasping calculus concepts.

A crucial aspect is to recognize that the function is undefined whenever the denominator equals zero; however, this is not a concern here as \( x^2 + 1 \) is always positive for real \( x \). Another point of interest with rational functions is how they asymptotically approach certain y-values as \( x \) increases or decreases.
  • Numerator: A polynomial \( x^2 - 1 \)
  • Denominator: A polynomial \( x^2 + 1 \)
  • No undefined points in real numbers as \( x^2 + 1 \gt 0 \)

Analyzing these elements provides insights into the nature of the function's graph and aids in understanding the roots, asymptotes, and other significant characteristics of the function.
Interval Computation
In calculus, intervals allow us to evaluate functions at a sequence of specific points to understand their behavior over a range. Here, we are given an interval from 0.5 to 10.5, increasing in steps of 1, as \( x = 0.5, 1.5, 2.5, \ldots, 10.5 \).

To compute the function over this interval, plug each \( x \) value into the function \( h(x) \) and calculate the result. This practice helps in visualizing the function behavior and identifying trends, such as increases, decreases, and leveling off.
  • Start at \( x = 0.5 \) and determine \( h(0.5) \).
  • Continue to \( x = 10.5 \) in increments of 1.
  • Observe the computed values: they can hint at how the function behaves holistically.

Evaluating these values provides a comprehensive viewpoint on how the rational function changes over this range, which is a crucial aspect of understanding its calculus properties such as limits and continuity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refer to the following graph, which shows the number \(f(t)\) of housing starts for single-family homes in the United States each year from 2000 through 2014 \((t=0\) represents \(2000,\) and \(f(t)\) is in thousands of units \():^{9}\) Estimate \(f(3), f(6),\) and \(f(8.5)\). Interpret your answers.

The production of ozone-layer damaging Freon 22 (chlorodifluoromethane) in developing countries rose from 200 tons in 2004 to a projected 590 tons in \(2010 .^{33}\) a. Use this information to find a linear model for the amount \(F\) of Freon 22 (in tons) as a function of time \(t\) in years since 2000 . b. Give the units of measurement and interpretation of the slope. c. Use the model from part (a) to estimate the 2008 figure and compare it with the actual projection of 400 tons.

The Snowtree cricket behaves in a rather interesting way: The rate at which it chirps depends linearly on the temperature. One summer evening you hear a cricket chirping at a rate of 140 chirps/minute, and you notice that the temperature is \(80^{\circ} \mathrm{F}\). Later in the evening the cricket has slowed down to 120 chirps/minute, and you notice that the temperature has dropped to \(75^{\circ} \mathrm{F}\). Express the temperature \(T\) as a function of the cricket's rate of chirping \(r\). What is the temperature if the cricket is chirping at a rate of 100 chirps/ minute?

Use correlation coefficients to determine which of the given sets of data is best fit by its associated regression line and which is fit worst. Is it a perfect fit for any of the data sets? a. \(\\{(1,3),(2,4),(5,6)\\}\) b. \(\\{(0,-1),(2,1),(3,4)\\}\) c. \(\\{(4,-3),(5,5),(0,0)\\}\)

The percentage of borrowers in the highest risk category who were delinquent on their payments decreased from \(9.7 \%\) in 2001 to \(4.3 \%\) in 2004 and then shot up to \(10.3 \%\) in \(2007 .^{40}\) Model this percentage \(P\) as a piecewise-linear function of the time \(t\) in years since 2001 , and use your model to estimate the percentage of delinquent borrowers in 2006 .

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.