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Question

The minimum value of f(x)=4x432x2+734(x24) for |x|>2, is

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Solution

Given : f(x)=4x432x2+734(x24)
Dividing numerator by denominator, f(x) reduced to :
f(x)=(x24)+94(x24)
and (x24)>0 for |x|>2,
using A.M.G.M.:
12[(x24)+94(x24)](x24)94(x24)(x24)+94(x24)2×32(x24)+94(x24)3
So, minimum value of f(x) is 3.

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