Find the solution of the inequality (1)/(|x|-3) ≤(1)/(2) (a) (-∞,-6] ∪(-2,2) ∪[6, ∞) (b) (-∞,-5]

Find the solution of the inequality (1)/(|x|-3) ≤(1)/(2) (a)...

Key Concepts

Absolute Value

Absolute value represents the distance of a number from zero, always as a non-negative quantity. In the context of inequalities, it is vital to consider how the expression inside the absolute value behaves, as it can affect the overall structure of the inequality, especially when combined with other operations, such as in rational expressions.

Domain Restrictions

When dealing with rational functions, it is crucial to identify values that cause the denominator to be zero. Excluding these values ensures the expression remains defined. In inequalities, finding the domain of the expression prevents including extraneous solutions that might arise from cases where the function is undefined.

Rational Inequalities

Rational inequalities involve fractions where both the numerator and denominator can contain variables. Solving them generally requires combining expressions onto a common denominator, isolating zero on one side, and then determining where the resulting expression is positive, negative, or zero. This process often involves testing intervals between critical points.

Critical Points Analysis

Critical points are values that make the numerator zero, the denominator zero, or otherwise change the behavior of the inequality. These points divide the number line into intervals over which the sign of the expression can be analyzed. Identifying these points is key to accurately solving inequalities as they mark where the function may change sign or become undefined.

Case Analysis in Inequalities

Case analysis is often required when solving inequalities that include absolute values or expressions where the sign of a component can change. By splitting the problem into distinct cases based on different ranges of the variable, each resulting inequality can be solved according to the sign constraints of that case. This ensures that all possible scenarios are considered and that the solution is accurate.

Transcript

00:01 Okay, so in this question we are given an expression which is 9 upon mod of x minus 5 minus 3.

00:12 It should be greater than equals to mode of x minus 2.

00:16 Okay, so we have two expressions of mode.

00:20 We know that whenever the questions come about mode, we have to see the critical points and we have to make the cases on the critical points.

00:29 So here we can see that critical points are x equals to 5 and x equals to 2.

00:36 Okay, so critical points are these.

00:39 Now we'll make cases upon 5 and 2.

00:42 Let's just take the first case, which is x is less than 2.

00:47 Then x will automatically be less than 5.

00:50 For x less than 2, our expression will be opened such as minus x plus 5 because x minus 5 will be a negative value then minus 3 is greater than equals to minus x plus 2.

01:07 Okay, now just simplifying this expression, x, sorry, 2 minus x is greater than 2 minus x.

01:20 Now 9 upon 2 minus x minus 2 minus x is greater than equals to 0.

01:27 Now note that x is less than 2 that means 2 minus x is a positive value okay so we'll just close multiply by 2 minus x just multiply by 2 minus x both side and it will not affect the inequality okay because 2 minus x is a positive value in this case so expression comes out to be 2 minus x squared is greater than 0 if we simplify it the quadratic expression comes out to be this one and we can just split it up like x minus 5 and x plus 1 is less than 0 and the range of x comes out to be minus 1 to 5 okay but note that x we have taken the case for x to be less than 2 okay so we have to take the intersection of x less than 2 and this one okay so final expression final result from this case comes out to be x belongs to minus 1 to 2 and 2 will be open bracket because x minus 2 should not be equals to 0 because it is in denominator okay so this is our first case right we have to take a union of all the we have to take union of all the cases okay now talking about case 2 case 2 will be x belonging to 2 to 5 okay for x belonging to 2 to 5 we have 9 upon x uh 2 minus x will be greater than equals to minus of 2 minus 6 okay now 9 upon 2 minus x plus 2 minus now note now note that here 2 minus x is a negative value because x is greater than 2 okay so when we will cross multiply by 2 minus x the inequality will be changed okay so the expression comes out to be 9 plus 4 plus x squared minus 4 x it should be less than equals to zero...

Please give Ace some feedback