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Prove the following inequalities: $\tan ^{-1}x < x$ for all $x>0$

Note: This is part 3rd of a 4 part question, split as 4 separate questions here.

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  • (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
  • (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
  • A function $f(x)$ is said to be increasing on an interval I $f(x_1) \leq f(x_2)$ whenever $x_1 < x_2$
  • It is strictly increasing if $ x_1 < x_2 =>f(x_1) < f(x_2)$
  • A function $f$ is said to be decreasing on an interval I if $f(x_1) \geq f(x_2)$ whenever $x_1 < x_2$
  • It is strictly decreasing if $ x_1 < x_2 =>f(x_1) > f(x_2)$
$\tan ^{-1} x < x $ for all $ x>0$
Let $f(x) =\tan ^{-1} x -x$
Step 1:
$f'(x)=\large\frac{1}{1+x^2}$$-1 \qquad x > 0$
for $ x >0 \large\frac{1}{1+x^2} <1 $
$\therefore f '(x) =\large\frac{1}{1+x^2}$$ -1 <0$
Step 2:
$f'(x) <0=>f(x)$ is strictly decreasing
Step 3:
$\therefore f(x) < f(0)$ for $ x > 0$
=>$ \tan ^{-1} x -x < 0 $ for $x > 0$
=>$ \tan ^{-1} x < x $ for $x > 0$
answered Jul 30, 2013 by meena.p
 
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