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Question
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) = `2x - 16/x^2`
and f"(x) = `2 + 32/x^3`
Consider, f'(x) = 0
∴ `2x - 16/x^2` = 0
∴ 2x = `16/x^2`
∴ x3 = 8
∴ x = 2
The maximum value is 2.
f(x) = `x^2 + 16/x`
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.
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