The real number x when added to its inverse gives the minimum sum at x equal :

Question

The real number $x$ when added to its inverse gives the minimum sum at $x$ equal :

Accepted Answer

$$f(x) = y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$

For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x =  \pm 1$$

$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $$

$$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$$  ($+ve$)

$$\therefore$$ f(x) will be minima at $x=1$.

Solutions · 1