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

Answers (1)

Given function is,
 f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }
f^{'}(x)\Rightarrow \frac{dy}{dx} = \frac{1}{1+x} - \frac{2 (2+x) - (1)(2x)} {(2+x)^{2} } = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^{2}}
                                                                                          = \frac{1}{1+x} - \frac{4}{(2+x)^2} = \frac{(2+x)^2 - 4(x+1)}{(x+1)(2+x)^{2}}
                                                                                          = \frac{4+x^{2} +4x -4x - 4}{(x+1)(2+x)^{2}} = \frac{x^{2} }{(x+1)(2+x)^{2}}
f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2}
Now, for  x > -1  , is is clear that   f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2} > 0  
Hence,   f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }  strictly increasing when  x > -1             

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