/*! 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 7 Which of the following determine... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 7

Which of the following determine a function \(f\) with formula \(y=f(x)\) ? For those that do, find \(f(x)\). Hint: Solve for \(y\) in terms of \(x\) and note that the definition of a function requires a single \(y\) for each \(x\). (a) \(x^{2}+y^{2}=1\) (b) \(x y+y+x=1, x \neq-1\) (c) \(x=\sqrt{2 y+1}\) (d) \(x=\frac{y}{y+1}\)

Short Answer

Expert verified
(b), (c), and (d) define functions as \(f(x) = \frac{1-x}{x+1}\), \(f(x) = \frac{x^2 - 1}{2}\), and \(f(x) = \frac{x}{1-x}\).

Step by step solution

01

Understanding Functions

A function is a relation where each input (domain element) corresponds to exactly one output (range element). For an equation to represent a function, solving for \(y\) must yield a unique value for each \(x\).
02

Analyzing Equation (a)

Consider the equation \(x^2 + y^2 = 1\). Solving for \(y\) gives \(y = \pm \sqrt{1 - x^2}\). Since there are two possible values of \(y\) for each \(x\), this equation does not define \(y\) as a function of \(x\).
03

Analyzing Equation (b)

For \(xy + y + x = 1\), rewrite as \(y = \frac{1 - x}{x + 1}\), assuming \(x eq -1\) to prevent division by zero. Each \(x\) in this modified domain corresponds to a unique \(y\), thus defining a function \(f(x) = \frac{1-x}{x+1}\).
04

Analyzing Equation (c)

Consider \(x = \sqrt{2y + 1}\). Solving for \(y\) gives \(y = \frac{x^2 - 1}{2}\). Since \(x\) maps to a single \(y\) value, this defines a function \(f(x) = \frac{x^2 - 1}{2}\).
05

Analyzing Equation (d)

Given \(x = \frac{y}{y+1}\), expressing \(y\) in terms of \(x\) leads to the quadratic \(yx + x = y\). Solving it gives \(y = \frac{x}{1-x}\) assuming \(x eq 1\) to prevent division by zero. Each \(x\) defines a unique \(y\), hence a function \(f(x) = \frac{x}{1-x}\).

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 91影视!

Key Concepts

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

Function Definition
A function is a special kind of relationship between two variables, usually `x` and `y`. In this relationship, each `x` value is linked to exactly one `y` value. It's like matching every student with one locker. This means that no matter what `x` you pick, there should be only one `y` value that pairs with it.
In mathematics, this is established by the vertical line test on a graph: if a vertical line crosses the graph of the equation at more than one point, then the relation is not a function. In algebraic terms, a function must be expressed with a clear and singular outcome for `y` when given `x`.
For example, if you have an expression like `y = 2x + 3`, for every `x` you input, you will get one and only one value of `y`.
Solving Equations
To figure out if an equation is a function, you often need to solve the equation for `y` in terms of `x`. This process involves rearranging and simplifying the equation until `y` stands alone on one side of the equation.
  • Start with the equation as given. For instance, if you have `x^2 + y^2 = 1`, your goal is to express `y` alone.
  • Apply algebraic techniques like factoring, expanding, or isolating the variable `y` depending on the structure of the equation.
  • The result should show only one value for every `x` value.
    For example, solving `xy + y + x = 1` gives `y = (1-x)/(x+1)`, which is a function because it gives a singular `y` for every `x` not equal to `-1`.
In cases where there are multiple possibilities for `y`, such as `y = +/鈭� sqrt(1-x^2)`, the relation does not define a function.
Unique Solutions
In the world of functions, unique solutions are key. This means for every `x`, there should be only one `y` paired with it. If multiple `y` values exist for a single `x`, it defies the function definition.
Consider `x = sqrt(2y + 1)`. Solving for `y` gives `y = (x^2 - 1)/2`. Here, for every `x` that you choose, you get only one corresponding `y`. This verifies that the equation represents a function.
Remember:
  • If solving the equation gives a square root or an absolute value, check carefully. These often produce more than one outcome and need reassessment.
  • For equations like `x = y/(y+1)`, ensure you handle denominators correctly to avoid errors in unique pairing.
Ensuring unique solutions involves careful algebraic manipulation to prevent multiple outcomes.
Domain and Range
The domain and range of a function describe where the function starts and ends on the graph. The domain is all the possible input values (or `x` values), while the range is all possible output values (`y` values). These aspects are crucial to correctly defining a function.
For an equation such as `y = (1-x)/(x+1)`, the domain is all numbers except `x = -1` (because this would cause division by zero). The range is all numbers that `y` can take when valid `x` values are plugged into the equation.
Understanding domains and ranges:
  • Make sure to exclude values that would result in division by zero from the domain.
  • Consider restrictions on square roots, as negative numbers aren't defined in the set of real numbers.
  • Think about the logical limits of `y` given the range of `x`.
Clearly defining these aspects ensures you have a functioning equation that behaves predictably across numbers it's applied to.

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

Use the addition identity for the tangent to show that \(\tan (t+\pi)=\tan t\) for all \(t\) in the domain of \(\tan t\).

Suppose that a tire on a truck has an outer radius of \(2.5\) feet. How many revolutions per minute does the tire make when the truck is traveling 60 miles per hour?

Find the length of the line segment joining the midpoints of the segments \(A B\) and \(C D\), where \(A=(1,3), B=(2,6)\), \(C=(4,7)\), and \(D=(3,4)\).

Calculate (be sure that your calculator is in radian or degree mode as needed). (a) \(\frac{56.4 \tan 34.2^{\circ}}{\sin 34.1^{\circ}}\) (b) \(\frac{5.34 \tan 21.3^{\circ}}{\sin 3.1^{\circ}+\cot 23.5^{\circ}}\) (c) \(\tan 0.452\) (d) \(\sin (-0.361)\)

Suppose that a continuous function is periodic with period 2 and is quadratic between \(-0.25\) and \(0.25\) and linear between \(-1.75\) and \(-0.25\). In addition, it has the value 0 at 0 and \(0.0625\) at \(\pm 0.25\). Sketch the function over the domain \([-2,2]\), and give a piecewise definition of the function.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation
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.