Write each solution set in interval notation. 4 x-x^3 ≥0 |...

Name: Problem 93, Chapter 1: Precalculus
Uploaded: 2021-05-01T01:16:35-08:00
Duration: 3 min 55 s
Channel: Kian Manafi
Description: Write each solution set in interval notation. 4 x-x^3 ≥0

Key Concepts

Interval Notation

Interval notation is a concise and clear method for describing sets of real numbers, especially those arising as solutions to inequalities. It uses specific symbols to denote whether endpoints are included (using brackets) or excluded (using parentheses) and helps in neatly presenting the solution set without ambiguity.

Critical Points and Sign Analysis

Critical points are the values that make each factor of a polynomial equal to zero. These points divide the number line into distinct intervals. Sign analysis involves picking test points within these intervals to determine whether the overall expression is positive, negative, or zero. This method is essential in solving inequalities, as it helps in understanding the behavior of the polynomial over different intervals.

Polynomial Inequalities

This concept involves solving inequalities that include polynomial expressions. It requires understanding where the polynomial is greater than, less than, or equal to zero. The process typically involves finding the zeros of the polynomial, which divides the number line into intervals that can be tested to determine where the inequality holds. It is a fundamental aspect of algebra that enhances problem-solving and analytical skills.

Factoring Polynomials

Factoring is the process of breaking down a polynomial into a product of simpler expressions (factors). This step is critical because it simplifies the polynomial and makes it easier to solve equations or inequalities by revealing the roots or critical points of the expression. Factoring transforms complex expressions into manageable pieces, facilitating further analysis like sign testing.

Transcript

00:01 We're trying to find a solution set to this interval or to this, sorry, to this inequality, and then put it in interval notation.

00:08 So the first thing i'm going to do is actually factor out in x.

00:12 So we're left with x times 4 minus x squared is greater than or equal to zero.

00:17 And now this stuff inside the parentheses is actually a difference of squares.

00:21 And if you're familiar with the difference of squares, what we can do is we can take the square root of the first term and then add the square root of the second term.

00:30 Multiply it by the square of the first term minus the square root of the second term.

00:35 And so now that we have our inequality in this form, what we can do is look at all the zeros of our inequality or where this side of our inequality is equal to zero.

00:45 So since we have this x multiplied by everything, we have a zero at x is equal to zero.

00:50 And since we have a factor of 2 plus x, we have x is also equal to negative 2 as a 0.

00:57 And 2 minus x means we have a factor at x is equal to the reason that i know we have a factor at negative 2 and 2 is if we plug in 2 into our equation this value goes to 0 and then we're multiplying two values by 0 which is always going to be equal to 0 and if we plug in negative 2 into this fact this factor of our inequality then we're going to get the same thing so now that we found these three zeros what we're going to do is we're going to look at three intervals from negative infinity or sorry four intervals from negative infinity to negative 2 from negative 2 to 0 from 0 to 2 and from 2 to infinity and we're going to look at what our equation sign is in these four intervals so for the first interval we have a value of x that's less than negative 2 so we're going to have we look at our equation again we're going to have a negative value for this first factor which is just x.

01:59 So we have a negative value times 2 plus x.

02:02 If we have an x that's less than negative then that's also going to be negative and 2 minus x is going to be positive since we're going to have 2 minus a negative number, which is going to be 2 plus a number.

02:14 So we're going to have a negative times a negative times a positive, which is equal to a positive value.

02:19 So this first interval is part of our solution set.

02:23 And now looking at the second interval, if we plug in a value of say negative 1, this value is going to be negative...

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