Solve each rational inequality by hand. Do not use a calculator. (1)/(x-1)+(1)/(x+1)>(3)/(4)

Name: Problem 70, Chapter 4: A Graphical Approach to College Algebra
Uploaded: 2020-07-20T12:02:16-08:00
Duration: 6 min 2 s
Channel: Ankit Gupta
Description: Solve each rational inequality by hand. Do not use a calculator. (1)/(x-1)+(1)/(x+1)>(3)/(4)

Solve each rational inequality by hand. Do not use a...

Key Concepts

Interval Testing and Sign Analysis

After determining the critical points, the number line is divided into intervals. Testing a sample value from each interval in the simplified inequality helps to determine whether the expression is positive or negative on that interval. This sign analysis is used to decide which intervals satisfy the original inequality.

Finding Critical Points

Critical points are the values where either the numerator or the denominator equals zero. These points divide the real number line into intervals and can be potential locations where the expression changes sign. Identifying these points is a vital step in analyzing and solving rational inequalities.

Rational Inequalities

Rational inequalities involve expressions that are ratios of two polynomials. Solving these inequalities requires finding the values of the variable that make the inequality true, taking into account the behavior of the numerator and denominator. It often involves finding a common denominator and then analyzing the sign changes in the resulting rational expression.

Domain Restrictions

In any rational expression, values that make the denominator zero are not allowed because division by zero is undefined. These restrictions must be identified at the start of the solving process and excluded from the solution set to ensure the answer is valid.

Common Denominator and Simplification

Combining rational expressions typically involves finding a common denominator, which allows different terms to be combined into a single fraction. This step is crucial as it simplifies the inequality, making subsequent steps such as combining like terms or factoring more straightforward.

Transcript

00:01 We have given with inequality 1 divided with x minus 1 plus 1 divided with x plus 1 is greater than 3 divided with 4.

00:10 Now send this term to the left of this.

00:15 So we will get left with 1 divided with x minus 1 plus 1 divided with x plus 1 minus 3 divided with 4 is greater 0.

00:24 Now we want this term as greater 0.

00:26 So that's why taking the here common.

00:30 Common x plus 1 so we left with x plus 1 plus x minus 3 multiplied with x minus 1 and x plus 1 is greater 0 so now we have to solve this so here 1 2 1 gets cancelled and we left with 2x minus 3 multiplied with x square so now this will be x and minus x gets cancelled then minus 1 divided with x minus 1 and x plus 1 greater 0 here also we have to solve this now we left with 3x square negative 3x square plus 2x plus 3 divided with x minus 1 and x plus 1 now we don't want a denominator as so this will be greater 0 we don't want denominator as 0 so that's why x minus 1 not equals to 0 and x plus 1 not equals to 0 here that means x not equals to 1 and x not equals to minus 1.

01:42 So domain is so domain of the function that is minus infinity to minus 1 in union 1 to infinity.

01:51 Here also one term missing that is that is minus 1 to 1 and then 1 to infinity.

01:58 And now we have to solve this inequality here minus 3 x squared plus 2x plus 3 divided with x minus 1 and x plus 1.

02:12 That is x square minus 1.

02:14 I can write it as minus 3 x square plus 2x plus 3 divided with x square minus 1.

02:21 Should be greater 0.

02:23 And now we have to solve the 0s of this inequality.

02:28 So for this, this can be written as minus 3 x square.

02:33 So here, this is not setting in the form of its roots.

02:37 So that's why i am writing the formula minus b plus minus here b squared minus 4ac divided with 2a.

02:48 Now get these values.

02:51 So now these are the value i can write it as minus 2 plus minus 2 here root 10.

02:58 And then minus 6 by more simplifying we get minus 1 plus minus root 10 divided with minus 3 now put 1 10 minus and then plus and solve we will we get alpha and beta are the two roots i am solving here so now by solving we get two roots here alpha and beta now put these values on number line here is your number line and this is your number line and now plot these point here so this is first minus 1 so here minus 1 and then here minus 720 and then here again 1 and then 1 .37 and then 1 .37 and here this is 1 .387 and here this is infinity and minus we have to take values.

04:01 So here also i am telling that minus 1 and 1 are holes and we want only positive sign.

04:10 So that's why we don't want zero values.

04:15 So these all will be the holes.

04:17 Now put any values between these intervals and find either it is positive or negative...

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