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Find the values of x for which the following func- tions are strictly increasing : f(x) = x3 – 6x2 – 36x + 7

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प्रश्न

Find the values of x for which the following func- tions are strictly increasing : f(x) = x3 – 6x2 – 36x + 7

बेरीज
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उत्तर

f(x) = x3 – 6x2 – 36x + 7

∴ f'(x) = `d/dx(x^3 - 6x^2 - 36x + 7)`

= 3x2 – 6 x 2x – 36 x 1 + 0
= 3x2 – 12x – 36
= 3(x2 – 4x – 12)
f is strictly increasing if f'(x) > 0
i.e. if 3(x2 – 4x – 12) > 0
i.e. if x2 – 4x –12 > 0
i.e.if x2 – 4x > 12
i.e. if x2 – 4x + 4 > 12 + 4
i.e. if (x – 2)2 > 16
i.e. if x – 2 > 4 or x – 2 < – 4
i.e if x > 6 or x < – 2
∴  f is strictly increasing if x < – 2 or x > 6.

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पाठ 2: Applications of Derivatives - Exercise 2.4 [पृष्ठ ८९]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 2 Applications of Derivatives
Exercise 2.4 | Q 2.3 | पृष्ठ ८९

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