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Question
Find the maximum and minimum of the following functions : f(x) = `x^2 + (16)/x^2`
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Solution
f(x) = `x^2 + (16)/x^2`
∴ f'(x) = `d/dx(x^2) + 16d/dx(x^-2)`
= 2x + 16(– 2)x–3
= `2x - (32)/x^3`
and
f"(x) = `d/dx(2x) - 32d/dx(x^-3)`
= 2 x 1 – 32(– 3)x–4
= `2 + (96)/x^4`
f'(x) = 0 gives `2x - (32)/x^3` = 0
∴ 2x4 – 32 = 0
∴ x4 = 16
∴ x = ± 2
∴ the roots of f'(x) = 0 are x1 = 2 and x2 = – 2
(a) f"(2) = `2 + (96)/(2)^4` = 8 > 0
∴ by the second derivative test, f has minimum at x = 2 and minimum value of f at x = 2
= f(2) = `(2)^2 + (16)/(2)^2`
= 4 + 4
= 8
(b) f"(– 2) = `2 + (96)/(-2)^4` = 8 > 0
∴ by the second derivative test, f has minimum at x = – 2 and minimum value of f at x = – 2
= f(– 2)
= `(- 2)^2 + (16)/(-2)^2`
= 4 + 4
= 8
Hence, the function f has minimum value 8 at x = ± 2.
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