y = x(x – 3)2 decreases for the values of x given by : ______.

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y = x(x – 3)2 decreases for the values of x given by : 1 < x < 3. Explanation: Here y = x(x – 3)2 `"dy"/"dx" = x * 2(x - 3) + (x - 3)^2 * 1 ⇒ "dy"/"dx" = 2x(x - 3) + (x - 3)^2` For increasing and decreasing "dy"/"dx" = 0 ∴ 2x(x – 3) + (x – 3)2 = 0 ⇒ (x – 3)(2x + x – 3) = 0 ⇒ (x – 3)(3x – 3) = 0 ⇒ 3(x – 3)(x – 1) = 0 ∴ x = 1, 3 ∴ Possible intervals are (– oo, 1), (1, 3), (3, oo) "dy"/"dx" = (x – 3)(x – 1) For (– oo, 1) = (–) (–) = (+) increasing For (1, 3) = (–) (+) = (–) decreasing For (3, oo) = (+) (+) = (+) increasing So the function decreases in (1, 3) or 1 < x < 3

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