Prove that, for all x>1 2 √(x)>3-(1)/(x)

Name: Problem 68, Chapter 3: Calculus
Uploaded: 2021-05-06T04:28:07-08:00
Duration: 5 min 48 s
Channel: Zoe Tillman
Description: Prove that, for all x>1 2 √(x)>3-(1)/(x)

Key Concepts

Domain Considerations

Domain considerations are fundamental to ensure that all operations performed during the manipulation of inequalities are mathematically valid. By clearly establishing that x > 1, one can safely apply operations such as division and square rooting without violating the conditions of the functions involved.

Squaring Both Sides

Squaring both sides of an inequality is a useful strategy to eliminate square roots, but it requires caution. The operation is only valid when both sides are known to be non-negative, as it can potentially introduce extraneous solutions. This concept is central when transforming an inequality to a form that is easier to solve.

Square Root Function and Domain

Understanding the square root function includes knowing that it is defined only for non-negative values, making it crucial to consider the variable's domain when working with it. In problems involving x > 1, this ensures that the square root is computed over a valid range and sets the stage for further manipulations.

Inequality Manipulation

Manipulating inequalities involves a set of algebraic techniques that maintain the inequality's truth. These techniques include adding, subtracting, multiplying, or dividing both sides by positive quantities, and taking valid operations such as squaring both sides when both expressions are non-negative, thereby simplifying the inequality for easier analysis.

Transcript

00:01 Well, everyone, so today we're going to take a look at this inequality and prove it to be true for all x greater than 1.

00:06 You might be asking yourself, what does this have anything to do with calculus, right? but as we go through, you'll see that it comes in handy near the end.

00:14 So first things first, i'm going to simplify this so it's easier for me to look at.

00:18 So what i'm going to do first is i'm going to square both sides to get rid of the root.

00:24 So 4x is greater than 3 minus 1 over x squared.

00:28 And this will equal, if you do it out, it'll be 9 minus 6 over x plus 1 over x squared.

00:41 Now i'm going to multiply both sides by x squared to get rid of this horrible denominator here.

00:49 We're going to have 4x cubed is greater than 9x squared minus 6x plus 1.

00:57 Now, one last thing, it might be nice if we add all our x terms on one side.

01:03 So i'm going to subtract 4x cubed from both sides.

01:09 0 will be greater than negative 4x cubed plus 9x squared minus 6x plus 1.

01:17 So, now semantically, what we're looking at is that we want to make sure that this thing is less than 0 for all x greater than 1.

01:30 So my plan is to observe at x equals 1 and see what's going on, and then observe whether the function decreases further from where it is from x equals 1.

01:50 Because we have to observe x equals 1 first, right? because if x equals 1 is positive, then whether it increases or decreases, it doesn't matter.

01:59 It's always going to be positive for a little bit greater than one, which isn't right.

02:04 But if it's negative, right, and it starts increasing, that's not good either.

02:09 So we're going to need to make sure that x is either negative, sorry, that y is either negative or zero at x equals 1, and then it continues to decrease for infinity, right? so let's give that a try.

02:25 First, let's do f of 1.

02:31 Equals right i'm just gonna should have said i'm just gonna call this f of x just to make life easier right so f of 1 is plus 9 minus 6 plus 1 so this evaluates to 0 okay so we have the case where this is our little x y axis thankfully x is not included in the inequality so that's good so we just need to make sure that that y decreases be beyond x equals 1.

03:06 Great.

03:07 And how do we do that? let's use our first derivative test and see where f prime of x is negative because where f prime of x is negative is going to be where f of x is decreasing...

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