Question

Show that y=log(1+x)-2x/2+x,x>-1, is an increasing function of x throughout its domain.

Answer 1

function is $\bf{y=log(1+x)-\frac{2x}{2+x},x>-1}$
differentiate y with respect to x ,
$\bf{\frac{dy}{dx}=\frac{1}{1+x}-\left[\begin{array}{c}\frac{(2+x)\frac{d}{dx}(2x)-2x\frac{d}{dx}(2+x)}{(2+x)^2}\end{array}\right]}\\\\=\bf{\frac{1}{1+x}-\frac{4+2x-2x}{(2+x)^2}}\\\\=\bf{\frac{1}{(1+x)}-\frac{4}{(2+x)^2}}$

dy/dx = 1/(1 + x) - 4/(2 + x)²
= {(2 + x)² - 4(1 + x)}/(1 + x)(2 + x)²
= {(x² + 4x + 4 - 4 - 4x}/(x + 1)(x + 2)²
=x²/(x + 1)(x + 2)²
hence, dy/dx = 1/(x + 1) {x/(x + 2)}²
here you see {x/(x + 2)}² is always positive for all real value of x
so, dy/dx depends upon 1/(x + 1)
if x < -1 then, dy/dx will be negative
if x > -1 then, dy/dx will be positive

but domain is x > -1 so, dy/dx > 0
therefore f is increasing throughout its domain