Answer to Question #283842 in Calculus for Tushal

Question

If x is positive show that. x>log(1+x)>x-(x^2/2)

Answer 1

To prove

$x>log(1+x)>x-\frac{x^{2}}{2}$ when $x>0$

We first suppose that

$f(x)=x-log(1+x)$

Then

$f^\prime(x)=1-\frac{1}{1+x}$

$\implies$ $f^\prime(x)=\frac{x}{1+x}$

$\implies$ $f^\prime(x)>0$ as $x>0$

$\implies$ $f(x)>f(0)$

$\implies$ $f(x)>0$ as $f(0)=0$

$\implies$ $x-log(1+x)>0$

$\implies$ $x>log(1+x) ----(1)$

Next, Suppose that

$g(x)=log(1+x)-(x-\frac{x^{2}}{2})$

when $x>0$

$\implies$ $g^\prime(x)=\frac{1}{1+x}-(1-x)$

$\implies$ $g^\prime(x)=\frac{1}{1+x}-1+x$

$\implies$ $g^\prime(x)=\frac{1-1-x}{1+x}+x$

$\implies$ $g^\prime(x)=x-\frac{x}{1+x}$

$\implies$ $g^\prime(x)=\frac{x^{2}}{1+x}$

$\implies$ $g^\prime(x)>0$ as $x>0$

So $g(x)$ is increasing for $x>0$

$\implies$ $g(x)>g(0)$

$\implies$ $g(x)>0$ as $g(0)=0$

$\implies$ $log(1+x)-(x-\frac{x^{2}}{2})>0$

when $x>0$

$\implies$ $log(1+x)>(x-\frac{x^{2}}{2})---(2)$

Combining (1) and (2), we have

$x>log(1+x)>x-\frac{x^{2}}{2}$

when $x>0$

Hence proved.