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Question:

Let f(x) = x² + 1/x² and g(x) = x - 1/x, x ∈ R - {-1, 0, 1}. If h(x) = f(x)g(x), then the local minimum value of h(x) is

2√2

3

√2

Solution:

We know that h(x) = f(x)g(x) = (x² + 1/x²)(x - 1/x) = (x - 1/x)² + 2(x - 1/x)
Applying AM ≥ GM we get (x - 1/x) + 2(x - 1/x) ≥ 2√2
Solving the above inequality, we get (x - 1/x)² + 2(x - 1/x) ≥ 2√2
Thus, the local minimum value of h(x) is 2√2