Finding the set of values for x

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In summary, the set of values for x which satisfies the inequality \frac{(x-3)^{2}}{x+1} < 2 is (-\infty, -1) \cup (1, 7).
  • #1
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Homework Statement


Find the set of values for x which

[tex]\frac{(x-3)^{2}}{x+1} < 2[/tex]

I turned this into

[tex]\frac{(x-7)(x-1)}{x+1} < 0 [/tex]

Now this implies that the critical angles are x = 7, 1 and -1 but how do I create the sets. I think I have to do somthing with sign change?

Thanks :)

EDIT: WHAT'S WRONG WITH MY LATEX??
 
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  • #2
There is nothing wrong with your latex; the forum is having problems at the moment.

It is a product of 3 terms, so you have the conditions abc < 0 iff (ab < 0 and c > 0) or (bc < 0 and a > 0) or (ac < 0 and b > 0). Inequalities usually generate logical sets like this.
 
  • #3
Hi thomas49th! :smile:
thomas49th said:
Find the set of values for x which

(x-3)2/(x+1) < 2

I turned this into

(x-7)(x-1)/(x+1) < 0

Now this implies that the critical angles are x = 7, 1 and -1 but how do I create the sets. I think I have to do somthing with sign change?

Yes, that's fine …

(and of course it's the same as the product (x-7)(x-1)(x+1) < 0 … do you see why? :wink:)

and you know that the sign can only change at 7 1 or -1 (but why are you calling them angles? :confused:), and will change each time (since there are no squared factors) :wink:
 
  • #4
Sorry I meant values NOT angles. My bad :S


It is a product of 3 terms, so you have the conditions abc < 0 iff (ab < 0 and c > 0) or (bc < 0 and a > 0) or (ac < 0 and b > 0). Inequalities usually generate logical sets like this.
Never done the sets like that. Pretty sure I've never been taught that method, though I can't remeber the method so that's why I am asking :)

(and of course it's the same as the product (x-7)(x-1)(x+1) < 0 … do you see why? )

and you know that the sign can only change at 7 1 or -1 (but why are you calling them angles? ), and will change each time (since there are no squared factors)

I don't know why you can multiply them like that. (x+1) is the denominator ! you could say (x+1)-1 maybe?

The way I say it outloud is:

y values of (x-7)(x-1) are divide by the y values x+1 graph and I want to find where the new composite graph is below the x-axis. I'm not sure what the most sensible way of doing it is... I am sure there was a method of drawing out a table?

Thanks :)
 
  • #5
thomas49th said:
I don't know why you can multiply them like that. (x+1) is the denominator ! you could say (x+1)-1 maybe?

Because (x+1)2 is always positive, so A/(x+1) has the same sign as (A/(x+1))(x+1)2, = A(x+1).
The way I say it outloud is …

Why are you making this so complicated?

Use slider142's :smile: method …

the product of three things will be negative if exactly an odd number of them is negative … ie if either one is negative, or all three are negative …

so in this case … ? :smile:
 
  • #6
sorted. Thanks :)
 

1. How do I find the set of values for x in an equation?

The set of values for x can be found by solving the equation for x. This can be done by using algebraic operations to isolate x on one side of the equation. Once x is isolated, the resulting expression will represent the set of values for x.

2. Can there be more than one set of values for x in an equation?

Yes, there can be more than one set of values for x in an equation. This is especially true for equations with variables such as exponents or radicals, where there could be multiple solutions for x.

3. Is it possible for the set of values for x to be infinite?

Yes, in some cases, the set of values for x can be infinite. This can occur when the equation has a variable in the denominator or when there are multiple solutions for x.

4. What should I do if I encounter an equation with no solutions for x?

If an equation has no solutions for x, it means that there is no value of x that satisfies the equation. In this case, the solution set for x would be empty, and there is no need to find a set of values for x.

5. Can I use a calculator to find the set of values for x?

While a calculator can be useful for solving equations, it is important to understand the steps involved in finding the set of values for x. Relying solely on a calculator may make it difficult to identify any errors or mistakes made in the solution process.

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