Inequality (1 Viewer)

[coyazayo]

Hi all! I have some questions on inequality i need help with. Thanks in advance

Graph the regions:

1. |xy|Greater than or equal to 1 (is x=0 included or excluded in this case?)

2.1/x>1/y

It would be appreciated if you showed working out as well

 

[Paradoxica]

Hi all! I have some questions on inequality i need help with. Thanks in advance

Graph the regions:

1. |xy|Greater than or equal to 1 (is x=0 included or excluded in this case?)

2.1/x>1/y

It would be appreciated if you showed working out as well

1. The inequality is the boundary defined by the plane curve x²y² = 1 and everything outside of it extending to infinity.

The infinitely long four point curved star formed by the inside region cannot satisfy this inequality.

2. Multiply both sides by x²y² to acquire y²x - x²y > 0

(xy)(y-x)>0

The region divides into six parts. Firstly, construct the line y=x.

Observe that (y-x) is positive in the region above the line y=x, and negative for the region below the line y=x

xy is positive in the first and third quadrants, and negative in the second and fourth quadrants.

So for y<x, the first and third quadrants are not valid, only the fourth quadrant satisfies the inequality.

and for y>x, the first and third quadrants are valid, whereas the second quadrant is invalid.

So the region is defined by the set of all points whose angular measure with respect to the origin is:

π/4<θ<π/2

-π/2<θ<0

-π<θ<5π/4

Excluding the lines y=x, x=0, y=0, since those make the expression impossible.

 

[InteGrand]

Hi all! I have some questions on inequality i need help with. Thanks in advance

Graph the regions:

1. |xy|Greater than or equal to 1 (is x=0 included or excluded in this case?)

2.1/x>1/y

1) Yes, exclude x = 0, because if x is 0, |xy| = 0, which isn't greater than or equal to 1.

The region in question must be symmetric about both coordinate axes. This is because if a point (x,y) satisfies the inequality, so does (+/- x, +/- y), for any independent choices of the +/-. So the region to shade can be found by finding the appropriate region in the first quadrant, and then just flipping about the axes and origin (so get the corresponding pictures in the other quadrants).

So when x, y > 0 (first quadrant), the inequation becomes xy >= 1, so this is the region on and above the hyperbola xy = 1 (which is y = 1/x) in the first quadrant.

So to get the full region, sketch in these hyperbolic branches in each quadrant and shade in the places on the sides of the branches that are "away" from the origin (just like in the first quadrant). The branches themselves are part of the region, since the inequation was ">=", rather than a strict ">".

2) Note that x and y are both non-zero, since otherwise there'll be a zero denominator. This means the coordinate axes are excluded from the region to shade. Note that any point (x,y) with x > 0 and y < 0 satisfies the inequation, so shade in the whole fourth quadrant.

Clearly no point in the second quadrant satisfies the inequation, since then the LHS (1/x) is negative whilst the RHS (1/y) is positive.

Now, when x and y have the same sign (first or third quadrant), we have 1/x > 1/y if and only if x < y. So in quadrants 1 and 3, shade in the places where y > x (i.e. above the line y = x in these quadrants).