Show that f(x)=2x+ ^ -1 x+ ( 1+x^1 -x ) is increasing in R. - Vidyadip

Question

Show that $\text{f(x)}=2\text{x}+\cot^{-1}\text{x}+\log\Big(\sqrt{1+\text{x}^1}-\text{x}\Big)$ is increasing in R.

✓

Answer

We have, $\text{f(x)}=2\text{x}+\cot^{-1}\text{x}+\log\Big(\sqrt{1+\text{x}^1}-\text{x}\Big)$
$\therefore\ \text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}+\frac{1}{\sqrt{1+\text{x}^2}-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}-1\Big)$
$=\frac{1}{1+\text{x}^2}+\frac{1}{\sqrt{1+\text{x}^2}-\text{x}}\cdot\frac{\text{x}-\sqrt{1+\text{x}^2}}{\sqrt{1+\text{x}^2}}=2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{1+\text{x}^2}}$
Now $1+\text{x}^2,\sqrt{1+\text{x}^2}\geq1$ for all real x.
$\therefore2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{1+\text{x}^2}}>0$ for all real x,
$\therefore\ \text{f(x)}>0$ for all real x.
Thus f(x) is increasing on R.

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