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

(i)

Rational inequalities

Without a calculator, solve 4x+3≤x\dfrac4{x+3}\le x.

Explore this idea

Hint

  1. State x≠−3x\ne-3. Move xx to the left and combine.

  2. Factorise the numerator.

  3. Use a sign chart.

Answer · (i)

−4≤x<−3-4\le x<-3 or x≥1x\ge1.

Explore the idea

Open interactive exploration

Where is (x + 4)(x − 1)/(x + 3) ≥ 0?

Test values in each interval.

(x+4)(x−1)x+3≥0\frac{(x+4)(x-1)}{x+3}\ge0

Choose an interval to test it. Shading marks the solution intervals.

Intervalx + 4x − 1x + 3product
negativenegativenegativenegative
positivenegativenegativepositive
positivenegativepositivenegative
positivepositivepositivepositive
Test value of xx = 2.00
Try an example

(x + 4)+(x − 1)+(x + 3)+⇒ product positive

Is it ≥ 0?
Yes
x is in the solution set

The shaded strips are where the expression satisfies the inequality; they match the green-tinted columns of the chart.

  • Expression, and where it is ≥ 0
  • Your test value
Key idea

Rewrite so the leading coefficient is positive and the sign chart alternates +, −, +, … from the right.

Ideas to try
Which intervals are closed and which are open at their ends?