Show that the function $f(x) = \log(1 + x) + \dfrac{1}{1 + x}$ increases on $(0, \infty)$.

> [!TLDR] > $f'(x)$ simplifies to $\dfrac{x}{(1+x)^2}$, which is positive for every $x>0$; hence $f$ increases on $(0,\infty)$. > [!FORMULA] > $f$ is increasing on an interval if $f'(x)>0$ throughout it. 1. Given $f(x)=\log(1+x)+\dfrac{1}{1+x}$. 2. Differentiate: $\dfrac{d}{dx}\log(1+x)=\dfrac{1}{1+x}$ and $\dfrac{d}{dx}(1+x)^{-1}=-\dfrac{1}{(1+x)^2}$. 3. So $f'(x)=\dfrac{1}{1+x}-\dfrac{1}{(1+x)^2}$. 4. Take LCD $(1+x)^2$: $f'(x)=\dfrac{(1+x)-1}{(1+x)^2}=\dfrac{x}{(1+x)^2}$. 5. For $x>0$: numerator $x>0$ and denominator $(1+x)^2>0$, so $f'(x)>0$. 6. Since $f'(x)>0$ for all $x\in(0,\infty)$, $f$ is (strictly) increasing on $(0,\infty)$. > [!ANSWER] > $f'(x)=\dfrac{x}{(1+x)^2}>0$ for all $x>0$; therefore $f$ increases on $(0,\infty)$. **Proved.**

Show that the function f(x) = log(1 + x) + (1)/(1 + x) increases on (0, ∞).

3.4 · Q3

f(x) = \log(1 + x) + \dfrac{1}{1 + x}

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Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

The Intuition

Imagine you are walking along a path that only ever goes uphill, or only ever goes downhill. You never have to go up and then down, or down and then up. That path is monotonic — it moves in one consistent direction.

A function is monotonic when its output (the $y$ -value) never reverses direction as its input (the $x$ -value) increases. If it always goes up (or stays flat), it is increasing. If it always goes down (or stays flat), it is decreasing. If it does both — rises, then falls — it is not monotonic.

Note

The word "monotonic" comes from Greek monotonos — "one tone." Just as a monotone voice stays on a single pitch, a monotonic function stays on a single trend.

The Precise Definition

Let $f$ be a function defined on an interval $I$ . We say:

$f$ is increasing (or non-decreasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \leq f(x_2)$ .
$f$ is strictly increasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) < f(x_2)$ .
$f$ is decreasing (or non-increasing) on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) \geq f(x_2)$ .
$f$ is strictly decreasing on $I$ if, for any $x_1 < x_2$ in $I$ , we have $f(x_1) > f(x_2)$ .

If a function is either increasing or decreasing on an interval, it is called monotonic on that interval.

Watch out

| Common Mistake | Correct Understanding |

|----------------|----------------------|

| "Increasing means $f'(x) > 0$ everywhere" | $f'(x) > 0$ implies strictly increasing, but a function can be increasing even where $f'(x) = 0$ at isolated points (e.g., $f(x) = x^3$ at $x=0$ ). |

| "Monotonic means the whole domain" | A function can be monotonic on a sub-interval but not on its entire domain. For example, $f(x) = x^2$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ , but not monotonic on $\mathbb{R}$ . |

How to Determine Monotonicity (The Derivative Test)

For differentiable functions, the derivative tells you the direction:

If $f'(x) \geq 0$ for all $x$ in an interval, then $f$ is increasing on that interval.
If $f'(x) \leq 0$ for all $x$ in an interval, then $f$ is decreasing on that interval.
If $f'(x) > 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly increasing.
If $f'(x) < 0$ for all $x$ in an interval (except possibly at isolated points), then $f$ is strictly decreasing.

f'(x) \geq 0 \implies f \text{ increasing} \qquad f'(x) \leq 0 \implies f \text{ decreasing}

Why It Matters

Monotonic functions are predictable. They have at most one root (if strictly monotonic), they are invertible (if strictly monotonic), and their extreme values occur only at the endpoints of the interval. In exam problems, you will often be asked to:

Find the intervals where a given function is increasing or decreasing.
Use monotonicity to prove inequalities (e.g., show $e^x > 1 + x$ for $x > 0$ ).
Determine the number of real roots of an equation.

A Worked Example

Problem: Find the intervals of monotonicity for $f(x) = x^3 - 3x + 1$ .

Solution:

First, find the derivative:

f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)

Set $f'(x) = 0$ to find critical points: $x = -1$ and $x = 1$ .

Now test the sign of $f'(x)$ in each interval:

Interval	Test point	$f'(x)$ sign	Conclusion
$(-\infty, -1)$	$x = -2$	$3(4-1) = 9 > 0$	Increasing
$(-1, 1)$	$x = 0$	$3(0-1) = -3 < 0$	Decreasing
$(1, \infty)$	$x = 2$	$3(4-1) = 9 > 0$	Increasing

Therefore, $f$ is increasing on $(-\infty, -1]$ and $[1, \infty)$ , and decreasing on $[-1, 1]$ .

Tip

| Always check the endpoints: if $f'(x) > 0$ on an open interval $(a,b)$ and $f$ is continuous at $a$ and $b$ , then $f$ is increasing on the closed interval $[a,b]$ . This is why we include the critical points in the answer above. |

The Big Picture

Monotonic function analysis is the study of trends. Instead of memorising formulas, think: "As I slide my finger to the right along the x-axis, does the graph go up, down, or stay flat?" The derivative is just a tool to answer that question precisely. Once you see the direction, you unlock the function's behaviour — roots, inverses, inequalities — all from a single, simple idea.

$f'(x)$ simplifies to $\dfrac{x}{(1+x)^2}$ , which is positive for every $x>0$ ; hence $f$ increases on $(0,\infty)$ .

f

Given $f(x)=\log(1+x)+\dfrac{1}{1+x}$ .
Differentiate: $\dfrac{d}{dx}\log(1+x)=\dfrac{1}{1+x}$ and $\dfrac{d}{dx}(1+x)^{-1}=-\dfrac{1}{(1+x)^2}$ .
So $f'(x)=\dfrac{1}{1+x}-\dfrac{1}{(1+x)^2}$ .
Take LCD $(1+x)^2$ : $f'(x)=\dfrac{(1+x)-1}{(1+x)^2}=\dfrac{x}{(1+x)^2}$ .
For $x>0$ : numerator $x>0$ and denominator $(1+x)^2>0$ , so $f'(x)>0$ .
Since $f'(x)>0$ for all $x\in(0,\infty)$ , $f$ is (strictly) increasing on $(0,\infty)$ .

✓Final answer

f'(x)=\dfrac{x}{(1+x)^2}>0

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