# Answer to Question #28760 in Integral Calculus for Ankit

Question #28760

-1/modulusx-2 >_1

Expert's answer

-1/|x|-2>=1

In interval notation we have the following:

________________0________________

So there are two intervals to be considered. Note that we exclude x=0 because we have x in the denominator, therefore it should be nonzero.

1)x<0

In this case we have

-1/(-x)-2>=1

1/x-2>=1

1/x>=3

As x<0 this inequality has no solutions so we have empty set

2)x>0

we obtain

-1/x-2>=1

-1/x>=3

1/x<=-3

1<=-3x

x<=-1/3

Also we have x>0 so there's also empty set

Final answer is empty set.

Also we could do it another way.

-1/|x|-2>=1

-1/|x|>=3

1/|x|<=-3

This is false because 1/|x|>0 for real x

In interval notation we have the following:

________________0________________

So there are two intervals to be considered. Note that we exclude x=0 because we have x in the denominator, therefore it should be nonzero.

1)x<0

In this case we have

-1/(-x)-2>=1

1/x-2>=1

1/x>=3

As x<0 this inequality has no solutions so we have empty set

2)x>0

we obtain

-1/x-2>=1

-1/x>=3

1/x<=-3

1<=-3x

x<=-1/3

Also we have x>0 so there's also empty set

Final answer is empty set.

Also we could do it another way.

-1/|x|-2>=1

-1/|x|>=3

1/|x|<=-3

This is false because 1/|x|>0 for real x

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