Question

 Use the sign of the derivative to establish the inequaltiy ln(1 + x) > x − x 2 2 , ∀ x > 0

Answer 1

Solution:-

Now , we solve with this assumption

$f(x) = ln(1+x) -x+x^2$

$\implies$

$f'(x ) = \dfrac{1}{1+x}-1+2x$

$\implies$

$f'(x) = \dfrac{x+2x^2}{1+x}$

$\implies$

For $x>0$ , We get $f'(x) >0$

Hence, we clearly say that $f(x)>f(0)$

Hence, $ln(1+x)-x+x^2>0$

=

$ln(1+x)>x-x^2$ (Proved).