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Question
Find the value of x for which the function f(x)= 2x3 – 9x2 + 12x + 2 is decreasing.
Given f(x) = 2x3 – 9x2 + 12x + 2
∴ f'(x) = `squarex^2 - square + square`
∴ f'(x) = `6(x - 1)(square)`
Now f'(x) < 0
∴ 6(x – 1)(x – 2) < 0
Since ab < 0 ⇔a < 0 and b < 0 or a > 0 and b < 0
Case 1: (x – 1) < 0 and (x – 2) < 0
∴ x < `square` and x > `square`
Which is contradiction
Case 2: x – 1 and x – 2 < 0
∴ x > `square` and x < `square`
1 < `square` < 2
f(x) is decreasing if and only if x ∈ `square`
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Solution
Given f(x) = 2x3 – 9x2 + 12x + 2
∴ f'(x) = 6x2 – 18x + 12 = 6(x2 – 3x + 2)
∴ f'(x) = 6(x - 1)(x – 2)
Now f'(x) < 0
∴ 6(x – 1)(x – 2) < 0
Since ab < 0 ⇔a < 0 and b < 0 or a > 0 and b < 0
Case 1: (x – 1) < 0 and (x – 2) < 0
∴ x < 1 and x > 2
Which is contradiction
Case 2: x – 1 and x – 2 < 0
∴ x > 1 and x < 2
∴ 1 < x < 2
∴ f(x) is decreasing if and only if x ∈ (1, 2)
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