A repeated factor does not disappear from a strict inequality

Show x2(x+2)2−1=(x+1)2(x+1+2)(x+1−2)x^2(x+2)^2-1=(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2). mark Hence solve 1/(x+2)2>x21/(x+2)^2>x^2 without a calculator. marks Deduce the inputs satisfying 1/(ex+2)2>e2x1/(e^x+2)^2>e^{2x}. marks

[modified 2008/RJC/Prelim/II/1]

Total marks: 6

Solution

(i)

Factorising

Show that x2(x+2)2−1=(x+1)2(x+1+2)(x+1−2)x^2(x+2)^2-1=(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2).

Hint

  1. Write x(x+2)=x2+2xx(x+2)=x^2+2x so the left side is a square minus 1.

  2. Difference of two squares twice.

Answer · (i)

x2(x+2)2−1=(x+1)2(x+1+2)(x+1−2)x^2(x+2)^2-1=(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2).

(ii)

Rational inequalities

Hence solve 1(x+2)2>x2\dfrac1{(x+2)^2}>x^2 without a calculator.

Explore this idea

Hint

  1. State what xx cannot be: x≠−2x\ne-2.

  2. (x+2)2>0(x+2)^2>0, so multiply by it safely.

  3. Reduce to the factorised quartic from (i).

Answer · (ii)

−1−2<x<−1+2-1-\sqrt2<x<-1+\sqrt2, excluding x=−2,−1x=-2,-1.

Explore the idea

A double factor and a double pole

Both (x + 1)² and (x + 2)² keep their sign, but the points x = −1 and x = −2 are still special.

(x+1)2(x+1+2)(x+1−2)(x+2)2<0\frac{(x+1)^2(x+1+\sqrt2)(x+1-\sqrt2)}{(x+2)^2}<0

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

Interval(x + 1)²x + 1 + √2x + 1 − √2(x + 2)²product
positivenegativenegativepositivepositive
positivepositivenegativepositivenegative
positivepositivenegativepositivenegative
positivepositivenegativepositivenegative
positivepositivepositivepositivepositive
Test value of xx = -1.50
Try an example

((x + 1)²)+(x + 1 + √2)+(x + 1 − √2)−((x + 2)²)+⇒ product negative

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 negative
  • Your test value
Key idea

Even-power factors keep the sign on both sides, but they carve a hole at their root. Three intervals result, not one.

Ideas to try
  1. Test x=−1x=-1 and x=−2x=-2. What goes wrong at each?
  2. How many separate intervals does the solution have?
(iii)

Exponential substitution

Deduce the values of xx satisfying 1(ex+2)2>e2x\dfrac1{(e^x+2)^2}>e^{2x}.

Explore this idea

Hint

  1. This is (ii) with xx replaced by exe^x.

  2. ex>0e^x>0, so only the positive part of the solution survives.

Answer · (iii)

x<ln⁡(2−1)x<\ln(\sqrt2-1).

Explore the idea

e^x below √2 − 1

Between y = 0 and y = √2 − 1 ≈ 0.414 the exponential curve lies to the left of x = ln(√2 − 1) ≈ −0.88.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Y = eˣ
  • Y = √2 − 1
  • Boundary
Ideas to try
Why does the lower bound −1−2-1-\sqrt2 not matter?