Question

Find all values of x for which `(x^3(x - 1))/((x - 2)) > 0`

Answer 1

The given inequality is f(x) = `(x^3(x - 1))/((x - 2)) > 0`

[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined.
When x = 2, f(x) = `oo` ⇒ f(x) is not defined.]

The critical numbers are x = 0, 1, 2

Divide the number line into 4 intervals

`(– oo, 0)`, (0, 1), (1, 2) and `(2, oo)`

(i) `(– oo, 0)`

When x < 0 say x = – 1

The factor x³ = (– 1)³ = – 1 < 0

The factor x – 1 = – 1 – 1 = – 2 < 0

The factor x – 2 = – 1 – 2 = – 3 < 0

∴ `(x^3(x - 1))/(x - 2) < 0`

Thus `(x^3(x - 1))/(x - 2) > 0` is not true in the interval `(– oo, 0)`

Therefore, it has no solution in the interval `(– oo, 0)`

(ii) (0, 1)

When 0 < x < 1 say x = 0.5

The factor x³ = (0.5)³ > 0

The factor x – 1 = 0.5 – 1 = – 0.5 < 0

The factor x – 2 = 0.5 – 2 = – 1.5 < 0

Thus x³ > 0, x – 1 < 0 and x – 2 < 0

∴ `(x^3(x - 1))/(x - 2) < 0`

Thus `(x^3(x - 1))/(x - 2) > 0` is true in the interval (0, 1)

Therefore, it has solution in (0, 1)

(iii) (1, 2)

When 1 < x < 2 say x = 1.5

The factor x³ = 0

The factor x – 1 = 1.5 – 1 = 0.5 > 0

The factor x – 2 = 1.5 – 2 = – 0.5 < 0

Thus x³ > 0, x – 1 > 0 and x – 2 < 0

∴ `(x^3(x - 1))/(x - 2) < 0`

Thus `(x^3(x - 1))/(x - 2) > 0` is not true in the interval (1, 2)

Therefore, it has no solution in (1, 2)

(iv) `(2, oo)`

When x > 2 say x = 3

The factor x³ = 3³ > 0

The factor x – 1 = 3 – 1 = 2 > 0

The factor x – 2 = 3 – 2 = 1 > 0

Thus x³ > 0, x – 1 > 0 and x – 2 > 0

∴ `(x^3(x - 1))/(x - 2) < 0`

Thus `(x^3(x - 1))/(x - 2) > 0` is true in the interval `(2, oo)`.

Therefore, it has a solution in `(2, oo)`.

Inteval	Sign of  x3	Sign of (x – 1)	Sign of (x – 2)	Sign of `(x^3(x - 1))/(x - 2)`
`(– oo, 0)`	–	–	–	–
(0, 1)	+	–	–	+
(1, 2)	+	+	–	–
`(2, oo)`	+	+	+	+

∴ `(x^3(x - 1))/(x - 2) < 0` has solution in the intervals (0, 1) and `(2, oo)`

∴ The solution set is given by (0, 1) ∪ `(2, oo)`