Suppose x lies in the interval (1, 3) with x≠2. Find the small...

Question

Suppose x lies in the interval (1, 3) with x≠2. Find the smallest positive value of δ such that the inequality 0<|x−2|<δ is true.

Accepted Answer

Hello. In this video, we are going to be finding the minimum positive value of the given inequality. So, we are told that if a variable W is selected from the interval of 0.5 to 2.5, with the condition that W is not equal to 1.5, we want to find the minimum positive value of lambda, such that the inequality, 0, less than the absolute value of W minus 1.5, less than lambda is satisfied. In order to approach this problem, let's go ahead and break down the given inequality. Now, the inequality, 0, less than the absolute value of minus 1.5, less than lambda can be rewritten as the absolute value of W minus 1.5 less than lambda. The reason why we omit the zero for now is because the zero indicates that the absolute value must be some positive value greater than 0. Now, we can go ahead and split up the inequality into two separate inequalities. We can rewrite the inequality as W minus 1.5 is less than lambda, and W minus 1.5 is greater than negative lambda. Now what we want to do is you want to go ahead and determine the value of W of the inequalities under the assumption that W is not equal to 1.5. So, on the left-hand side, we will add 1.5 to both sides of the inequality, leaving us with W less than lambda plus 1.5. Repeating this process on the right hand side allows us to get a value of W greater than 1.5 minus lambda. So, what this is saying is that W must be between the values of lambda plus 1.5, and lambda. A 1.5 minus lambda. So what we need to do now is we need to determine the bounds of W. In order to do this, we can go ahead and take the boundary points of the interval given to us. Now, the interval given to us is 0.5.2.5. Let's go ahead and plug in the first boundary, 0.5 into the original inequality. That will give us the absolute value of 0.5 minus 1.5. This is going to give us the output of one, and this is going to be less than lambda. Next, if we take the boundary 0.2.5, and plug it into the original inequality, we will have 2.5 minus 1.5, which is equal to 1, and this will be less than lambda. So what this is saying is that the value of lambda that makes the inequalities above true, must have a minimum positive value of 1. So in other words, lambda must equal to one, which is going to be the solution to this problem. And with that being said, the overall solution is going to bed. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Suppose x lies in the interval (1, 3) with x≠2. Find the smallest positive value of δ such that the inequality 0<|x−2|<δ is true.

Key Concepts

Absolute Value Inequality

Intervals

Limit and Continuity

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