Maharashtra State BoardHSC Commerce 12th Board Exam
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Determine the maximum and minimum value of the following function. f(x) = x2+16x - Mathematics and Statistics

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Sum

Determine the maximum and minimum value of the following function.

f(x) = `"x"^2 + 16/"x"`

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Solution

f(x) = `"x"^2 + 16/"x"`

∴ f'(x) =`2"x" - 16/"x"^2`

and f''(x) = `2 + 32/"x"^2`

Consider, f'(x) = 0

∴ `2"x" - 16/"x"^2 = 0`

∴ 2x = `16/"x"^2`

∴ x3 = 8

∴ x = 2

For x = 2

`f''(2) = 2 + 32/2^3 = 2 + 32/8 = 2 + 4 = 6 > 0`

∴ f(x) attains minimum value at x = 2.

∴ Minimum value = `"f"(2) = (2)^2 + 16/2 = 4 + 8 = 12`

∴ The function f(x) has minimum value 12 at x = 2.

Concept: Maxima and Minima
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Applications of Derivatives
Exercise 4.3 | Q 1.3 | Page 109
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