Solve the inequality 1/x<or=[x-2] for real numbers?

1 Answer
Jun 20, 2018

Assuming no brackets, 12x<0 and x1+2

Explanation:

It is a little unclear from the question exactly what is meant - the brackets are redundant. I'll supply answers for two possible question cases - a) Brackets removed; b) Brackets changed to modulus brackets.

a) Brackets removed

1xx2

Recall that multiplying through by a negative number implies changing the direction of an inequality, so:

If x0, then 1x(x2), so x22x10
If x<0, then 1x(x2), so x22x10

Solve the quadratic for x22x1=0 by the quadratic formula:

x=2±4+42=1±2

Then we know that the quadratic expression will change whether it satisfies the inequality or not at these points. Evaluate it at some sample points to check whether it is positive or negative on these intervals:

Let f(x)=x22x1. Then for the first interval (x<12), f(1)=1+21=2, so f(x) is positive in this interval. For the second interval (12<x<1+2), f(0)=001=1, so f(x) is negative in this interval. For the third interval, f(3)=961=2, so f(x) is again positive.

Compare these to the piecewise pair of inequalities that we are solving:

The first interval (x<12) lies entirely within the negative x range, so we want x22x10. But f(x) is positive on this interval, so there are no solutions.

The second interval (12<x<1+2) spans x=0, so the needed inequality changes in the middle of the interval. f(x) is negative on this interval, so we fulfil our equation when we are at x<0, but not when x>0. Thus the range 12x0 is one solution of the inequality.

The third interval (1+2<x) lies entirely within the positive x range, so we want x22x10. We noted that x is positive on this interval, so x1+2 is a second solution of the inequality.

In summary - two solution ranges for x:
12x<0 and x1+2
Note that the expression is not valid at x=0, due to the presence of the 1x term.

b) Brackets changed to modulus brackets

1x|x2|

There are two cases here that we didn't have to worry about above:

If x2, then 1xx2
If x<2, then 1x(x2)

Then we proceed as before, but combining these extra ranges into our working:

If 2x, then 1x(x2), so x22x10
If 0x<2, then 1x(x2), so x2+2x10
If x<0, then 1x(x2), so x2+2x10

We now have three ranges rather than the two we had before. One of them has the same quadratic expression as we had before; the other two have a similar expression with one coefficient sign change.

Solve the second quadratic for x2+2x1=0 by factorisation:

(x1)2=0x=1

This has two repeated roots at x=1. Consider the sign of the function g(x)=x2+2x1 - as it has only one root, it touches the x axis without passing through it - so it has the same sign over the whole function (excepting the single zero value) - evaluating it at e.g. 0 shows us quickly that that sign is negative.

Consider our three piecewise inequalities above.

When x<0, we need x2+2x10. We've established that the quadratic expression here is always negative or zero, so every x such that x<0 is a solution.

When 0x<2, we need x2+2x10. Only one point in this range fulfils this inequality: x=1, the zero point of g(x).

When 2x, we need x22x10, the original quadratic expression, f(x). Recall that the roots of f(x) are at x=1±2. The higher of these is approximately 2.414, which is >2, so we will have a change in inequality satisfaction at this point too. From our answer above, we see that f(x)0 when x1+2, and that it is negative between 2 and this point. So x1+2 is a third answer to this problem.

In summary - three solution ranges to this problem, one of which is a single point:
x<0, x=1, and x1+2
Note as before that the inequality is undefined at x=0 due to the presence of the 1x term.

Compare the solution cases between the two bracket possibilities. The solutions that give positive terms on both sides of the inequality are identical, as we would expect; otherwise they differ.