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

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Aicharya · Accepted Answer

$y=x+\frac{1}{x} 	ext { or } \frac{d y}{d x}=1-\frac{1}{x^2}$ For $\max$. or $\min ., 1-\frac{1}{x^2}=0 \Rightarrow x=\pm 1$ $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{2}{\mathrm{x}^3} \Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}ight)_{\mathrm{x}=2}=2(+$ ve minima) Therefore $x=1$

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

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