Rational Inequalities

Summary

This video tutorial offers a step-by-step guide on solving rational inequalities. It begins with an example, illustrating how to set up a number line, identify critical points, and determine the sign of each region. The video explains how to interpret the inequality 'greater than or equal to zero' and 'less than zero', and how to express the solution in interval notation and as inequalities. It also covers a method to solve inequalities involving non-zero constants by rearranging the inequality and finding common denominators. The tutorial concludes with a suggestion to explore more videos on the topic in the precalculus playlist.

Takeaways

• 📚 The video focuses on solving rational inequalities, starting with the example \( \frac{x - 3}{x + 2} \geq 0 \).
• 📏 A number line is used to identify critical points where the numerator or denominator equals zero, which are -2 and 3.
• ⭕️ At -2, an open circle is used on the number line because the function is undefined there, while at 3, a closed circle indicates the numerator is zero.
• 🔍 Three regions are identified on the number line to determine the sign of the expression in each region.
• 🤔 Testing points within each region helps determine if the expression is positive or negative, which is essential for solving inequalities.
• 📈 The solution to the first example is expressed in interval notation as \( (-\infty, -2) \cup [3, \infty) \), and can also be described using inequalities.
• 🔄 The process is repeated for a second example, \( \frac{(x - 4)(x + 1)}{x - 3} < 0 \), with points of interest at -1, 3, and 4.
• 🚫 For the second example, all points of interest are open circles since the inequality is strict (less than zero).
• 📉 The solution for the second example is given as \( (-\infty, -1) \cup (3, 4) \), and can be described using inequalities as well.
• 🔄 A third example involves moving a constant to the other side of the inequality and finding a common denominator to simplify the expression.
• 📌 The third example's solution is \( (-\infty, 1) \cup [2.5, \infty) \), with specific attention to the points where the expression equals zero or is undefined.
• 👍 The video concludes with a suggestion to check out the precalculus playlist for more videos on similar topics.

Q & A

What is the main topic of the video?

The main topic of the video is solving rational inequalities.

What is the first example of a rational inequality given in the video?

The first example is \( \frac{x - 3}{x + 2} \geq 0 \).

How does the video suggest to start solving the first example?

The video suggests starting by setting the numerator and denominator equal to zero to find points of interest on the number line.

What are the points of interest for the first example and why are they significant?

The points of interest are -2 and 3. They are significant because they are the values that make the numerator or denominator zero, which helps in determining where the function is undefined or where the sign of the expression changes.

Why is there an open circle at -2 on the number line for the first example?

There is an open circle at -2 because the denominator is zero at this point, making the function undefined, so it should not be included in the solution set.

How does the video determine the sign of the expression in different regions of the number line?

The video picks test points in each region, substitutes them into the expression, and checks whether the result is positive or negative.

What is the solution to the first example in interval notation?

The solution is \( (-\infty, -2) \cup [3, \infty) \), with an open parenthesis at -2 and a closed bracket at 3, indicating the inclusion or exclusion of these points.

What is the second example of a rational inequality presented in the video?

The second example is \( \frac{(x - 4)(x + 1)}{x - 3} < 0 \).

How does the video handle the inequality sign in the second example?

The video focuses on finding regions where the expression is negative, as the inequality is less than zero.

What is the final solution to the second example using interval notation?

The solution is \( (-\infty, -1) \cup (3, 4) \), indicating the intervals where the expression is negative.

What adjustment does the video make to the third example before solving it?

The video moves the constant '3' to the other side of the inequality and combines the terms over a common denominator before solving.

How does the video suggest expressing the solution to the third example as inequalities?

The solution can be expressed as \( x < -1 \) or \( x \geq \frac{5}{2} \), which corresponds to the regions where the expression is less than or equal to zero.

What is the advice given at the end of the video for finding more resources on the topic?

The video suggests checking out the precalculus playlist for more videos on similar topics.

Outlines

00:00

📚 Solving Rational Inequalities

This paragraph introduces the process of solving rational inequalities with a specific example: (x - 3) / (x + 2) ≥ 0. The explanation begins by setting up a number line, identifying critical points where the numerator or denominator equals zero, and determining where the function is undefined (x + 2 ≠ 0). The solution involves testing the sign of the expression in different regions of the number line and shading the regions where the expression is positive, as this is required by the inequality. The final solution is presented in interval notation and as an inequality, showing two different ways to express the answer.

05:02

🔍 Further Exploration of Rational Inequalities

The second paragraph continues the theme of solving rational inequalities, presenting a new example: (x - 4)(x + 1) / (x - 3) < 0. The process involves identifying points of interest and determining the sign of the expression in various regions. The solution requires finding where the expression is negative, as indicated by the inequality. The explanation includes the importance of multiplicity in zeros and how it affects the sign of the expression. The final solution is given in interval notation and as an inequality, providing clarity on the regions where the inequality holds true.

Keywords

💡Rational Inequalities

Rational inequalities involve expressions with rational functions, which are fractions where both the numerator and the denominator are polynomials. In the context of the video, solving these inequalities means finding the values of the variable for which the inequality holds true. The video script focuses on how to solve such inequalities by using a number line and testing regions for the sign of the rational expression.

💡Number Line

A number line is a visual tool used to represent real numbers in a linear, ordered format. In the script, the number line is used to plot points of interest derived from the rational inequality, which helps in determining the intervals where the inequality is satisfied. The number line is essential for visualizing the solution to the inequality.

💡Numerator

The numerator is the top part of a fraction. In the context of solving rational inequalities, setting the numerator equal to zero helps identify critical points that may affect the solution set. The script mentions setting the numerator equal to zero to find points where the rational expression could change signs.

💡Denominator

The denominator is the bottom part of a fraction and is crucial in rational expressions because it cannot be zero, as division by zero is undefined. The script discusses setting the denominator equal to zero to find points that must be excluded from the solution set, as they make the expression undefined.

💡Open Circle

An open circle on a number line indicates that a particular value is not included in the solution set. In the script, open circles are used at points where the denominator of the rational expression is zero, as these points make the expression undefined and therefore cannot be part of the solution.

💡Closed Circle

A closed circle on a number line signifies that a particular value is included in the solution set. The script uses closed circles at points where the numerator is zero, as these points are part of the solution when the inequality is satisfied by the rational expression being zero.

💡Sign of the Region

The sign of a region refers to whether the value of the rational expression is positive or negative within that region of the number line. The script describes testing the sign of the rational expression in different regions to determine where the inequality holds true.

💡Interval Notation

Interval notation is a way to express the solution set of an inequality in a concise form, using parentheses and brackets to indicate whether endpoints are included or excluded. The script provides examples of interval notation to express the solution sets for the rational inequalities discussed.

💡Inequality

An inequality is a mathematical statement that compares expressions, indicating that one is greater than, less than, or equal to another. In the script, inequalities are the focus, with the goal of finding the values of the variable that satisfy the given inequality conditions.

💡Multiplicity

Multiplicity refers to the number of times a factor appears in a polynomial. In the context of the script, the multiplicity of a zero affects how the signs of the rational expression alternate across different regions on the number line.

💡Common Denominators

Common denominators are necessary when combining or comparing fractions with different denominators. The script mentions finding a common denominator to simplify the rational inequality into a single fraction, which is then used to solve the inequality.

Highlights

Introduction to solving rational inequalities with a step-by-step example.

Setting up the number line with points of interest for the inequality x - 3 / (x + 2) ≥ 0.

Identifying critical points where the numerator or denominator is zero, resulting in undefined or zero values.

Using test points to determine the sign of the function in different regions of the number line.

Shading regions where the fraction is positive to find the solution set for the inequality.

Expressing the solution in interval notation and as inequalities.

Approach to solving a second example with the inequality (x - 4)(x + 1) / (x - 3) < 0.

Marking open circles at points of interest for less than zero inequalities.

Checking the signs in different regions to find where the expression is negative.

Determining the solution using interval notation for the second example.

Solving a third example with the inequality (x + 2) / (x - 1) ≤ 3 by rearranging the terms.

Creating a common denominator to simplify the inequality to a single fraction.

Identifying new critical points and determining the sign of the function in different regions for the third example.

Shading the region where the inequality is less than or equal to zero for the third example.

Expressing the solution for the third example in interval notation and as inequalities.

Encouragement to check out the precalculus playlist for more videos on related topics.

Closing remarks and thanks for watching the video on solving rational inequalities.

Transcripts

in this video we're going to focus on

solving rational inequalities

so let's start with this example x minus

3 divided by x plus 2

is greater than or equal to 0.

so let's start with a number line

now we need to set the numerator

equal to zero

so x is equal to three

and if we set the denominator equal to

zero

the second point of interest is negative

two so let's put that in the number line

negative two and three

to the right we have positive infinity

to the left negative infinity

notice that we have x plus two in the

bottom

and if you have a zero on the bottom the

function is undefined

so x plus two should not equal zero

therefore at negative two we should have

an open circle

but the numerator can be equal to zero

so

we're gonna have a closed circle

at three make sure you understand that

now

we need this fraction to be greater than

zero that means it has to be a positive

number

there's three regions that are important

in this number line

these three regions

we need to determine the sign of each of

those regions

so let's start with the first one

let's plug in 4

into this fraction

4 minus 3

is positive

and four plus two is positive

a positive number divided by a positive

number will give us a positive result

now let's pick a number between negative

two and three let's try zero

so zero minus three is negative

0 plus 2 is positive

a negative divided by a positive number

will give us a negative result

and if we try a number in this region

like negative 3

negative 3 minus 3

that's negative 6 so that's negative

negative three plus two is negative

and two negative numbers divided against

each other

will produce a positive result

now we want the fraction to be greater

than or equal to zero so we want the

positive values

so we're going to shade that region

and the region on the left

so the solution in interval notation

is going to be negative infinity to

negative 2

with a parenthesis due to the open

circle

and then union

3 to infinity using brackets due to the

closed circle that we have here

now to express the same answer using

inequalities

for this part you could say that

x

is less than a negative 2

or

for this part

x

is equal to or greater than 3.

so those are the two ways in which you

can express your answer

so that's how you can solve

this particular rational inequality

let's try another example

x minus four

times x plus one

divided by x minus three

and this is going to be less than zero

feel free to pause the video if you want

to

so the points of interest are negative

one

and then three

and then the next one is four

we're going to have positive infinity

and negative infinity

now this one is in the denominator so

we're going to have an open circle

at three

now notice that it doesn't equal zero

it's less than zero so in fact all of

these points of interest

will be open circle

if it was less than or equal to zero

these two

would be a closed circle

so keep that in mind

now

because it's less than zero

we want to find the regions that are

negative not positive

so let's check the signs

let's pick a number between four and

infinity let's use five

so five minus four

that's going to be

positive

and 5

plus 1 is positive

5 minus 3 is positive

so the first result will be positive

now if we pick a number between 3 and 4

like 3.5

3.5 minus 4 is negative

3.5 plus 1 positive

and 3.5 minus 3 is positive

so a negative times a positive times a

positive will give us a negative number

now

picking a number between negative one

and three

let's try zero zero minus four is

negative zero plus one is positive zero

minus three is negative

a negative times a positive times a

negative

will give us a positive number

now the signs will always alternate if

the multiplicity of each zero is one

so the next one is going to be negative

if you plug in negative two

this will be negative negative two minus

four is negative negative two plus one

is negative and negative two minus three

is negative

three negatives multiplied to each other

will be negative

now we want the portion of the graph

that is less than zero or that's

negative

so the answer using interval notation is

going to be negative infinity

to negative one union

three to four

as an inequality we could say that x is

less than negative one

or

x is between

four between three and four

and that's it for this problem

let's work on this one x plus two over x

minus one

let's say it's less than or equal to

three

so what do you think we need to do in

this case

because this is not zero

so we need to do some work before we can

make a number line let's move the 3 to

the other side

so it's going to be x plus 2

over x minus 1

minus 3.

and so that's less than or equal to zero

now

what we need to do at this point

is multiply negative three by x minus

one

over x minus one

we need to get common denominators

so this is gonna be x plus two

minus three times x minus one

and because the denominators are the

same we can write it as one single

fraction

so now let's distribute the negative

three to x minus one

so it's gonna be negative three x plus

three

and now let's simplify

so x minus three x

is negative two x

and two plus three

is five

so this is what we now have

now

we know that the denominator cannot be

zero so therefore x can't be one

and if we set the numerator equal to

zero

then five is equal to two x

so x is five over 2 or we could say 2.5

so those are the points of interest

so we have 1

and 2.5

now we're going to have a closed circle

at 2.5

and an open circle at one because you

can't have a zero in the bottom

now let's test the points

so let's try a number that's bigger than

two point five let's try three

so negative two times three plus five

over three minus one

negative two times three is six i mean

negative six plus five

that's negative

and three minus one is positive

so

here we have a negative result in this

region

and if we pick a number between one and

two point five like two

negative two times two plus five

over

two minus one it's negative four plus

five

that's positive two minus one is

positive

so this is going to be positive in that

region

and then for the last part

between negative infinity and 1 let's

try 0.

so if we try 0

this is just going to be 5 and this is

going to be negative 1

which will give us a negative result

so we want the answers that are less

than 0

so we want the negative answers

so we're going to shade the blue region

so the answer is going to be negative

infinity to 1 but not including 1

and then 5 over 2

to infinity

so as an inequality x is less than one

or

x is equal to or greater than five over

two

and that's it

if you want to find more of my videos on

these topics

check out my precalculus playlist you

might find other videos that

may be helpful so thanks again for

watching and have a good day

you