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

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

Explanation: We are given: \[ y = \log (1 + x) - \dfrac{2x}{2 + x} \] Differentiate with respect to $x$: \[ \dfrac{dy}{dx} = \dfrac{1}{1 + x} - \dfrac{(2 + x)(2) - 2x(1)}{(2 + x)^2} = \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2} \] Take LCM and simplify: \[ \dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \] Now, consider the domain $x > -1$. Let us examine the sign of the derivative: \[\] The numerator $x^2 \geq 0$, and is 0 only when $x = 0$ \[\] The denominator $(1 + x)(2 + x)^2 > 0$ for all $x > -1$, since all terms are positive. Therefore, \[ \dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \geq 0 \text{ for all } x > -1, \] $\textbf{Conclusion:} $ Since $\dfrac{dy}{dx} > 0$ for all $x > -1$, the function is $\textbf{increasing throughout its domain}$.

Question

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

Accepted Answer

Explanation: We are given:  
$$
y = \log (1 + x) - \dfrac{2x}{2 + x}
$$

Differentiate with respect to $x$:  
$$
\dfrac{dy}{dx} = \dfrac{1}{1 + x} - \dfrac{(2 + x)(2) - 2x(1)}{(2 + x)^2}
= \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2}
$$

Take LCM and simplify:  
$$
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2}
$$

Now, consider the domain $x > -1$. Let us examine the sign of the derivative:
 The numerator $x^2 \geq 0$, and is 0 only when $x = 0$
 The denominator $(1 + x)(2 + x)^2 > 0$ for all $x > -1$, since all terms are positive.

Therefore,  
$$
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \geq 0 \text{ for all } x > -1,
$$

$\textbf{Conclusion:} $ 
Since $\dfrac{dy}{dx} > 0$ for all $x > -1$, the function is $\textbf{increasing throughout its domain}$.

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