Reciprocals reverse the order of negative intervals

(i) Without a calculator solve 4/(x+3)≤x4/(x+3)\le x. marks

(ii) Hence solve 4x/(1+3x)≤1/x4x/(1+3x)\le1/x. marks

[2014/ACJC/Promo/2]

Total marks: 6

Solution

If this part uses an earlier result, revisit (i) Rational inequalities or open all working together.

(ii)

Reciprocal substitution

Hence solve 4x1+3x≤1x\dfrac{4x}{1+3x}\le\dfrac1x.

Explore this idea

Hint

  1. Divide the numerator and denominator of 4x1+3x\frac{4x}{1+3x} by xx to see 1x\frac1x.

  2. Replace xx by 1x\frac1x in the answer to (i).

  3. Invert each interval carefully: the sign of xx decides the direction.

Answer · (ii)

−1/3<x≤−1/4-1/3<x\le-1/4 or 0<x≤10<x\le1.

Explore the idea

Open interactive exploration

Reverse intervals through a reciprocal

Move the upper end of the negative output interval. Its input endpoints reverse order; poles stay excluded.

Original condition: −4 ≤ 1/x < −3 or 1/x ≥ 1. The original fractions exclude x = 0 and x = −1/3. Reset restores the question; changing the boundary explores a different output condition.

Negative interval upper bound-3

Open endpoint: the boundary is excluded.

Try an example

Required output values

[-4, -3) ∪ [1, ∞)

Required output values-606

Retained inputs

(-0.333333, -0.25] ∪ (0, 1]

Retained inputs-1.201.2

Reached outputs

[-4, -3) ∪ [1, ∞)

Reached outputs-606

Original allowed inputs

(−∞, -0.333333) ∪ (-0.333333, 0) ∪ (0, ∞)

These exclusions stay in force when a boundary changes. Arrows indicate continuation beyond the drawing window.

On each side of zero, reciprocation reverses endpoint order and carries each endpoint’s inclusion with it. Neither original denominator zero can be restored by closing an output boundary.

The graph shows a finite window; arrows and interval notation describe the full set. Decimal endpoints are rounded.