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Chapter 2: Problem 25

Identify the set of values \(x\) for which \(y\) will be a real number. $$ y=\frac{2}{x-3} $$

Short Answer

Expert verified
x can be any real number except 3.

Step by step solution

01

Identify Denominator Restrictions

To determine when the fraction is undefined, identify values that make the denominator zero. The function becomes undefined when the denominator is zero because division by zero is undefined.
02

Solve the Denominator Equation

Set the denominator equal to zero and solve for the variable. For the given equation, set \(x-3 = 0\).
03

Solve for x

Solve the equation \(x - 3 = 0\). Adding 3 to both sides, we get \(x = 3\). This value for \(x\) makes the denominator zero.
04

Determine Valid x Values

To ensure that \(y\) is a real number, exclude any values of \(x\) that make the denominator zero. Therefore, \(x\) can take any real value except 3.

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Key Concepts

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

undefined denominator
When analyzing rational functions, identifying when the denominator is undefined is crucial. A function becomes undefined when you try to divide by zero. Imagine dividing a cake. If you have zero pieces, you can't give any cake to anyone.

In our exercise, we have the rational function:
$$ y=\frac{2}{x-3} $$

To find when this function is undefined, set the denominator to zero:
$$ x-3 = 0 $$
This means that our function becomes undefined when:
$$ x = 3 $$
Any other value of x is acceptable and will make the function defined.
real numbers
Real numbers include all positive and negative integers, fractions, and decimals---practically any number you can think of. Imaginary numbers are excluded. In our scenario, we are interested in excluding the values that make the function undefined, keeping real numbers in consideration.

For our function:
$$ y=\frac{2}{x-3} $$
We acknowledge that x can be any real number except 3, as discussed. Numbers like 4, -2.5, or even 100 are all acceptable values for x, as long as we don't include the excluded values.
exclusion of values
To ensure that the function y remains a real number, we need to exclude specific values from x. This step helps us avoid making the denominator zero, which would make the function undefined. In our example,
$$ y=\frac{2}{x-3} $$
the only value that makes the denominator zero is 3, so we exclude it. The set of all possible x values (the domain) is thus all real numbers except for 3. Formally, we write this as:
\(x eq 3\)


Remember, excluding these critical values ensures the function produces real, understandable, and consistent results.
solving equations
Solving equations is a systematic approach that helps us identify critical values and understand functions better. To solve for the values where the function y is undefined:
  • Start by identifying the denominator.
  • Set the denominator equal to zero.
  • Solve for the variable.
For our function:
$$ y=\frac{2}{x-3} $$
We set the denominator equal to zero:
$$ x-3 = 0 $$
Solve for x by adding 3 to both sides:
$$ x = 3 $$
This value is then excluded from the possible x values. Thus, by solving such simple equations, we can better understand the behavior of functions and ensure they remain well-defined.

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Most popular questions from this chapter

Evaluate the function for the given values of \(x\). \(f(x)=\left\\{\begin{aligned}-3 x+7 & \text { for } x<-1 \\ x^{2}+3 & \text { for }-1 \leq x<4 \\ 5 & \text { for } x \geq 4 \end{aligned}\right.\) a. \(f(3)\) b. \(f(-2)\) c. \(f(-1)\) d. \(f(4)\) e. \(f(5)\)

Refer to the functions \(r, p,\) and \(q .\) Evaluate the function and write the domain in interval notation. \(r(x)=-3 x \quad p(x)=x^{2}+3 x \quad q(x)=\sqrt{1-x}\) $$(r-p)(x)$$

Graph the function. $$ s(x)=\left\\{\begin{array}{ll} -x-1 & \text { for } x \leq-1 \\ \sqrt{x+1} & \text { for } x>-1 \end{array}\right. $$

Given \(m(x)=\sqrt{x-4},\) evaluate \((m \circ m)(x)\) and write the domain in interval notation.

A car accelerates from 0 to \(60 \mathrm{mph}(88 \mathrm{ft} / \mathrm{sec})\) in \(8.8 \mathrm{sec} .\) The distance \(d(t)\) (in \(\mathrm{ft}\) ) that the car travels \(t\) seconds after motion begins is given by \(d(t)=5 t^{2},\) where \(0 \leq t \leq 8.8\) a. Find the difference quotient \(\frac{d(t+h)-d(t)}{h}\). Use the difference quotient to determine the average rate of speed on the following intervals for \(t:\) b. [0,2] c. [2,4] d. [4,6] e. [6,8]
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