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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)

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प्रश्न

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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उत्तर

Given f(x) = 2x3 – 9x2 + 12x + 2

∴ f'(x) = 6x218x + 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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