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Find the values of x such that f(x) = 2x^{3} – 15x^{2} + 36x + 1 is increasing function

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#### Solution

f(x) = 2x^{3} – 15x^{2} + 36x + 1

∴ f'(x) = 6x^{2} – 30x + 36

= 6(x^{2} – 5x + 6)

= 6(x – 3)(x – 2)

f(x) is an increasing function, if f'(x) > 0

∴ 6(x – 3)(x – 2) > 0

∴ (x – 3) (x – 2) > 0

ab > 0 ⇔ a > 0 and b > 0 or a < 0 and b < 0

∴ Either (x – 3) > 0 and (x – 2) > 0

or

(x – 3) < 0 and (x – 2) < 0

**Case 1:** x – 3 > 0 and x – 2 > 0

∴ x > 3 and x > 2

∴ x > 3

**Case 2:** x – 3 < 0 and x – 2 < 0

∴ x < 3 and x < 2

∴ x < 2

Thus, f(x) is an increasing function for x < 2 or x > 3 ,i.e., (– ∞, 2) ∪ (3, ∞)

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