INEQUALITIES HELP!! IMPORTANTOOO (1 Viewer)

[kingkong123]

HEY GUYS, HOW DO U SOLVE THIS INEQUALITY?

 

[Shadowdude]

edit: nevermind me.

 

[Sanical]

Ah, well, assuming x is positive?
Since 1/x is easy to sketch, just sketch the hyperbola and find where the hyperbola is less than 1 but greater than -1. You'll end up getting x>1, x<-1 (include the equal sign in the inequality)

 

[nightweaver066]

Graph the components. i.e. y = 1, y = -1, y = 1/x

Obviously your graph isn't going to look as perfect as this, but assume this is a rough sketch.

Now using y = 1/x, find where y = 1 cuts it.
When y = 1, 1 = 1/x, x = 1.

Now we need to find where y = -1 cuts it.
When y = -1, -1 = 1/x, x = -1.

From the graph, by inspection, the solution is x>=1, x<=-1 (this region of the graph is in between y = 1 and y = -1).

 

[Sanical]

Where's the latex button gone I can't find it.

 

[nightweaver066]

Where's the latex button gone I can't find it.

It's gone. If you want to use latex, type

 

[Sanical]

Ah wtf. So I need to know how to type in latex? Why did it go

 

[nightweaver066]

Ah wtf. So I need to know how to type in latex? Why did it go

Not too sure why it's gone, it went with the forum update..

You can use this latex editor, then insert whatever you type there in to the

 

[Sanical]

Not too sure why it's gone, it went with the forum update..

You can use this latex editor, then insert whatever you type there in to the

 

[kingkong123]

Graph the components. i.e. y = 1, y = -1, y = 1/x

Obviously your graph isn't going to look as perfect as this, but assume this is a rough sketch.

Now using y = 1/x, find where y = 1 cuts it.
When y = 1, 1 = 1/x, x = 1.

Now we need to find where y = -1 cuts it.
When y = -1, -1 = 1/x, x = -1.

From the graph, by inspection, the solution is x>=1, x<=-1 (this region of the graph is in between y = 1 and y = -1).

thanks heaps! but i was looking for a quicker way, not really solving graphically, is there an easier option?

 

[Drongoski]

Nightweaver's graphic approach is still the simplest and most intuitive method.

Algebraically, you can multiply the inequality by x^2 as per the usual method, (noting that x must be nonzero) and solve for the pair of implied quadratic inequalities:

 

[kingkong123]

Nightweaver's graphic approach is still the simplest and most intuitive method.

Algebraically, you can multiply the inequality by x^2 as per the usual method, (noting that x must be nonzero) and solve for the pair of implied quadratic inequalities:

when you solve those two quadratic ineuqalities you get x<=-1 , x>=0 and x<=0 , x>=1 respectively. but x =/=0 so you would cancel out the equality for the x>=0 and <=0, but wouldn't the combined range for the above solutions be x is all real excluding x=0???

thanks

 

[nightweaver066]

thanks heaps! but i was looking for a quicker way, not really solving graphically, is there an easier option?

Solving it graphically is already pretty quick if you know your graphs well and can simply reproduce them.

You could rewrite the inequality like,
and

and solve.

I'll take the first one,





Now solving the second part,







Now by combining these two results, we find that and contradict with the other solutions so they are not apart of it.

So then the solution is also x >=1, x <= -1.

 

[Drongoski]

when you solve those two quadratic ineuqalities you get x<=-1 , x>=0 and x<=0 , x>=1 respectively. but x =/=0 so you would cancel out the equality for the x>=0 and <=0, but wouldn't the combined range for the above solutions be x is all real excluding x=0???

thanks

In this case "x = 0" doesn't come into play.

But it is a good habit when solving for inequalities involving the unknown in the denominator to ensure we take into account & to exclude (whenever it becomes necessary) the cases of the denominator being 0.