Solve each rational inequality by hand. (1)/(x-1)+(1)/(x+1)>(3)/(4)

Name: Problem 70, Chapter 5: A Graphical Approach to Precalculus with Limits
Uploaded: 2020-08-18T22:48:44-08:00
Duration: 5 min 26 s
Channel: Amy Jiang
Description: Solve each rational inequality by hand. (1)/(x-1)+(1)/(x+1)>(3)/(4)

step 1 Okay, so we want to solve the following rational inequality. So let's start by moving all of our

Solve each rational inequality by hand....

Key Concepts

Interval Testing

After determining the critical points, the next step is to divide the number line into intervals, and then select test points within those intervals to evaluate the sign of the expression. This process, often called sign chart analysis, helps in determining where the inequality is satisfied. It is a systematic approach to ensure that all possibilities have been considered in a rational inequality.

Rational Inequalities

A rational inequality is an inequality that involves rational functions, that is, expressions that are ratios of polynomials. Solving these inequalities typically requires finding critical points—where either the numerator or denominator is zero—and determining where the inequality holds true by analyzing the sign of the expression in the intervals defined by these points.

Critical Points

Critical points are the values that make either the numerator or denominator of the rational expression equal to zero. These points are essential because they often serve as boundaries between intervals where the inequality may change its sign. Identifying these points allows one to construct a sign chart to test the inequality across different intervals.

Domain Restrictions

This concept involves identifying the values for which the rational expressions are undefined. For any rational expression, the denominator cannot equal zero, which means that certain values must be excluded from the solution set. Recognizing these restrictions is crucial when solving inequalities, as they determine the valid domain over which the solution can exist.

Combining Rational Expressions

When dealing with multiple rational expressions added or subtracted in a problem, it is often necessary to combine them into a single expression by finding a common denominator. This process simplifies the inequality and prepares it for further algebraic manipulation, such as moving all terms to one side or simplifying complex fractions.

Transcript

00:02 Okay, so we want to solve the following rational inequality.

00:05 So let's start by moving all of our terms to our left -hand side.

00:09 So here we're going to have a minus 3 over 4 is greater than 0.

00:15 Okay, so let's see, let's see, let's multiply this by x plus 1 over x plus 1.

00:26 I want to combine this into a single faction.

00:28 Here i'm going to multiply by an x minus 1 over x minus 1.

00:32 And here i'll multiply by an x plus 1 times x minus 1.

00:41 Okay, so here we're going to have an x plus 1.

00:47 Oh, we're also going to need to multiply by a 4.

00:51 Since we have a 4 here, but actually, it's this rewrite this as 3 over 4 in our numerated then.

01:04 So we have x plus 1 over x minus 1, and this is going to be plus x minus 1 minus 1 minus 3 over 4.

01:12 And this is x plus 1 times x minus 1 which is equal to x squared minus 1.

01:21 Okay, so we have minus 3 over 4, x squared, and then plus a 3 over 4.

01:34 Okay, so we currently have the pollen.

01:37 And now let's simplify our numerator.

01:39 So this is going to give us a minus 3 over 4, x squared, and then we have plus 2x.

01:50 And we have 1 minus 1, and then that's just 3 over 4, all over x minus 1 times x plus 1.

02:02 So can we factor out our numerator? okay, so yes, we can.

02:06 We can factor this into, let's factor out a negative 1 over 4.

02:11 And then we have 3x plus 1 and a x minus 3.

02:21 This is all over x minus 1 times x plus 1.

02:25 So we want to know when this is greater than zero.

02:28 So here, let's plot out our, or let's draw out our number line here and plug in our when our numerator and our denominator is equal to zero.

02:37 So that's going to be at negative 1 over 3, at 3, and at 1, as well as negative 1.

02:46 Okay, so let's note that we're going to have open circles at negative 1 and 1, since if we have x is equal to negative 1 or 1, we have division by 0...

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