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Prove the following inequalities: $\log(1+ x)< x $ for all $x>0$

Note: This is part 4th 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)$
$\log (1+x) < x$ for all $x > 0$
Let $f(x) =\log (1+x) -x$
Step 1:
$f'(x)=\large\frac{1}{1+x^2} $$-1$
Step 2:
$\large\frac{1}{1+x^2} < 1$ for $ x > 0$
$\therefore f'(x) < 0 $ for $ x > 0$
=>$f(x) $ is a strictly decreasing function , $ x > 0$
Step 3:
$\therefore f(x) < f(0)$
$\log (1+x) -x < \log 1-0$
$\log (1+x) -x <0$
=> $\log (1+x) < x$ for $ x < 0$
answered Jul 30, 2013 by meena.p
 

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