Question

Show that $y = log(1+x) - \frac {2x}{2+x}, \ x>-1$,is an increasing function of x throughout its domain.

Answer 1

We have,

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

$\frac {dy}{dx}$ = $\frac {1}{1+x}$ - $\frac {(2+x)(2)-2x(1)}{(2+x)^2}$ = $\frac {1}{1+x}$ -$\frac {4}{(2+x)^2}$ = $\frac {x^2}{(2+x)^2}$

Now, $\frac {dy}{dx}$ = 0

$\frac {x^2}{(2+x)^2}$ = 0

x2=0 [(2+x)≠0 as x>-1]

x=0

Since x >−1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1<x<0 and x>0. When −1<x<0, we have:

x<0$\implies$x2>0

x>-1$\implies$(2+x)>0 = (2+x)2>0

y' = $\frac {x^2}{(2+x)}$>0

Also, when x > 0:

x>0$\implies$x2>0, (2+x)2>0

$\implies$y'=$\frac {x^2}{(2+x)^2}$>0

Hence, function f is increasing throughout this domain.