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Question
Find the values of x for `y = [x(x - 2)]^2` is an increasing function.
Solution
We know, y = [x (x – 2)]² = x² (x + 4 – 4x)
= x4 – 4x3 + 4x2
On differentiating with respect to x,
`dy/dx` = 4x2 - 12x2 + 8x
= 4x (x2 - 3x + 2)
= 4x ( x - 1) (x - 2)
`dy/dx` = 0
`=>` 4x ( x - 1) (x - 2) = 0
`therefore` x = 0, 1, 2
∴ Four parts of real number line from x = 0, x = 1, x = 2 are intervals.
(`- infty`, 0), (0, 1), (1, 2), (2, 2) are formed.
| Interval | (∞, 0) | (0, 1) | (1, 2) | (2, ∞) |
| Sign of x | -ve | +ve | +ve | +ve |
| sign of (x - 1) | -ve | - ve | +ve | +ve |
| sign of (x - 2) | -ve | - ve | -ve | +ve |
| sign of `dy/dx` | -ve | +ve | -ve | +ve |
| nature of function | decreasing | increasing | decreasing | increasing |
∴ y is an increasing function in (0, 1) ∪ (2, ∞)
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