Let f(x) = x2 + \frac{1}{x^{2}}and g(x) = x – \frac{1}{x}, x \in R – {– 1, 0, 1}. If h(x) = \frac{f \left(\right. x \left.\right)}{g \left(\right. x \left.\right)}, then the local minimum value of h (x) is :

Engineering

Mathematics

Maxima and Minima

Question

Let f(x) = x² + $\frac{1}{x^{2}}$ and g(x) = x – $\frac{1}{x}$ , x $\in$ R – {– 1, 0, 1}. If h(x) = $\frac{f (x)}{g (x)}$ , then the local minimum value of h (x) is :

–3

$- 2 \sqrt{2}$

$2 \sqrt{2}$

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h(x) = $\frac{f (x)}{g (x)} = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}}$ x – $\frac{1}{x} = t$

$= \frac{t^{2} + 2}{t}$

h(x) = t + $\frac{2}{t}$

$h (x) \geq 2 \sqrt{2}$

Point of local minimum value is $2 \sqrt{2}$ .