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.