Exercise 6.2 · Q7: $\log(1+x)-\frac{2x}{2+x}$ is increasing for $x>-1$

Basics you need

Solution

1. The domain is exactly $x > -1$

   $\log(1 + x)$ needs $1 + x > 0$⁠, that is $x > -1$⁠. The fraction needs $2 + x \ne 0$⁠; for $x > -1$ we have $2 + x > 1 > 0$⁠, so this is automatic. So the domain is the interval $(-1, \infty)$⁠, and on it $y$ is differentiable (hence continuous).

   "Throughout its domain" means on the whole interval $(-1, \infty)$⁠. Knowing that $1 + x > 0$ and $2 + x > 0$ there is also what will settle the sign of the derivative in Step 4.
   The logarithm is defined only for positive numbers (Section 5.4). Basics: interval notation

2. Differentiate each piece

   First piece, chain rule with inner function $1 + x$⁠:

   $$\frac{d}{dx}\log(1 + x) = \frac{1}{1 + x}.$$

   Second piece, quotient rule with $u = 2x$⁠, $v = 2 + x$⁠, $u' = 2$⁠, $v' = 1$⁠:

   $$\begin{aligned} \frac{d}{dx}\left(\frac{2x}{2 + x}\right) &= \frac{2(2 + x) - 2x\cdot 1}{(2 + x)^2}\\ &= \frac{4}{(2 + x)^2}. \end{aligned}$$
   In the quotient rule the $2x$ terms cancel, leaving the constant 4 on top. That cancellation is what makes the final answer so clean.
   Quotient rule: $\frac{u'v - uv'}{v^2}$⁠; the order in the numerator matters. Basics: the chain rule

3. Combine over a common denominator

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{1 + x} - \frac{4}{(2 + x)^2}\\ &= \frac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2}\\ &= \frac{4 + 4x + x^2 - 4 - 4x}{(1 + x)(2 + x)^2}\\ &= \frac{x^2}{(1 + x)(2 + x)^2}. \end{aligned}$$
   To compare the two terms we need one fraction. The common denominator is $(1 + x)(2 + x)^2$⁠. Expanding $(2 + x)^2 = 4 + 4x + x^2$⁠, the constants and the $x$ terms cancel, leaving just $x^2$⁠.
   $(a + b)^2 = a^2 + 2ab + b^2$⁠, so $(2 + x)^2 = 4 + 4x + x^2$⁠.

4. Sign of the derivative for $x > -1$

   • Top: $x^2 \ge 0$⁠, and $x^2 = 0$ only at $x = 0$⁠.
   • Bottom: $1 + x > 0$ and $(2 + x)^2 > 0$ (Step 1), so the bottom is positive.

   Hence

   $$\frac{dy}{dx} \ge 0 \text{ for all } x > -1,$$

   with $\frac{dy}{dx} = 0$ only at $x = 0$ and $\frac{dy}{dx} > 0$ everywhere else in the domain.

   The single zero of the derivative does not change its sign: it comes from a square. So there is no turning point, only a momentary flat spot at $x = 0$⁠.
   Positive over positive is positive; a square is never negative. Basics: sign of a product

5. Conclude with Theorem 1

   Take any $x_1 < x_2$ in $(-1, \infty)$⁠. On $[x_1, x_2]$⁠, $y$ is continuous, it is differentiable on $(x_1, x_2)$⁠, and $\frac{dy}{dx} \ge 0$ there. By Theorem 1 (a), $y$ is increasing on $[x_1, x_2]$⁠, so $y(x_1) \le y(x_2)$⁠. As $x_1 < x_2$ were arbitrary, $y$ is an increasing function of $x$ throughout its domain.

   The theorem is stated on a closed interval, while the domain is the open, unbounded interval $(-1, \infty)$⁠. Applying it on $[x_1, x_2]$ for every pair inside the domain covers the whole domain.

6. Stronger: $y$ is strictly increasing

   On $(-1, 0)$ and on $(0, \infty)$ the derivative is strictly positive. Since $y$ is continuous at 0, the proof of Theorem 1 on $[x_1, x_2]$ gives strict increase on $(-1, 0]$ and on $[0, \infty)$⁠. For a pair on opposite sides, $x_1 < 0 < x_2$⁠, we get $y(x_1) < y(0) < y(x_2)$⁠. So $y$ is in fact strictly increasing on $(-1, \infty)$⁠.

   The question only asks for "increasing", which Step 5 proved. This step shows that the flat spot at $x = 0$ does not even produce two equal values.

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