Question

Find the set of values of x such that

Answer 1

\(x>\frac{6}{x}+1\)

Answer 2

@Zepdrix

Answer 3

Subtracting some stuff,

\(\large\rm x-\frac6x-1>0\)

Factoring out a 1/x,

\(\large\rm \frac1x(x^2-x-6)>0\)

Factoring further,

\(\large\rm \frac1x(x-3)(x+2)>0\)

Answer 4

Notice that we have equality, not greater than, at exactly x=3 and x=-2.

So let's see what's happening around those points.

Answer 5

okay

Answer 6

We know that the function is exactly zero at 3 and -2.

Answer 7

wow my picture got erased, whuuut

Answer 8

xD

Answer 9

Ok

Answer 10

So hmm let's see what's happening at x=1.
We really only care about the SIGN of the result because that's how we can compare it to zero.

Answer 11

looks like it's going to be negative overall, right?
Do you understand what I did there?

Answer 12

yes

Answer 13

I don't care about the actual number, only the sign.

Answer 14

a negative result is NOT going to be greater than 0.
So this middle interval does NOT satisfy our inequality.

Answer 15

You can look back at Dan's notes if you would like another approach.
He rewrote it as a quadratic and drew the shape of it.

Just another approach.

Answer 16

So then test the left interval over there,

Answer 17

The method u use is almost similar with my book.
yeap,Dan's method is also good.

I got (+) when I plug in x=-3

Answer 18

And (+) > 0

good good good.

You should get the same result when you test a point greater than 3.

Answer 19

Just remember that you're dealing with a STRICT inequality,
so you won't be able to include that end point.

Answer 20

how do you get (+) when you put in a -3?

Answer 21

Mmmm uh oh >.< my bad

Answer 22

I insert x=-3
\((x-3)(x+2)\)
\((-3-3)(-3+2)\)
\( (-)(-)=(+)\)

Answer 23

what happened to the 1/x?

Answer 24

Oh boy I forgot about this important location x=0.

Answer 25

0 needs to be on the number line too

Answer 26

Yes, ty for that correction.

Answer 27

I c. xD

Answer 28

Ya let's not forget about that thing in front. :)

Answer 29

Ok

Answer 30

I should've looked at the graph before answering your question :) lol
Got myself a little mixed up there.

Answer 31

So we have a couple more locations to check out, ya?

Answer 32

yes

Answer 33

x=-1, \((+)>0\)

Answer 34

I had one of my factors mixed up, woops.

Answer 35

x=2, \((-)<0\)

Answer 36

We didn't really need x=2, we already took care of that interval.
But that's ok.

Answer 37

ok

Answer 38

So this interval is part of your solution.
Ok great.

Answer 39

so,the answer would be
\(-2<x<0 or x>3\)

Answer 40

\({-2<x<0~or~x>3}\)

Answer 41

Yes.... I think so.

Answer 42

Hmm weird problem :)

Answer 43

okay,thanks @Zepdrix @Zarkon