Question

Show that y is an increasing function throughout its domain.

Answer 1

\[\LARGE y=\log(1+x)-\frac{2x}{2+x},x>-1\]

Answer 2

Differentiating,
\[\LARGE \frac{1}{1+x}-\frac{-2(2+x)-2x(1)}{(2+x)^2}=0\]
\[\LARGE \frac{1}{1+x}=\frac{-2(2+x)-2x(1)}{(2+x)^2}\]

Answer 3

I am completely def to algebra .. is this for my question ?

Answer 4

@ganeshie8 :D

Answer 5

thank you..

Answer 6

to do this 9 = z +6 they say I have to choose a transformation.. I am not sure of this.

Answer 7

\(\LARGE y=\log(1+x)-\frac{2x}{2+x},x>-1 \)

differiating

\(\LARGE \frac{dy}{dx} =\frac{1}{1+x}+\frac{-2x-2(2+x)}{(2+x)^2},x>-1 \)

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

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

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

Answer 8

Using @rsadhvika 's result, we see that

$$
\dfrac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}\\
=\dfrac{x^2+4x+4-4-4x}{(1+x)(2+x)^2}\\
=\dfrac{x^2}{(1+x)(2+x)^2}
$$
And since \( x>-1\),\(x^2\ge 0\implies x\ge 0\). So y is increasing.