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