To solve the inequality \( \log(1 + x) \leq x \), we will follow these steps:
### Step 1: Define the function
Let \( f(x) = \log(1 + x) - x \). We want to find the values of \( x \) for which \( f(x) \leq 0 \).
### Step 2: Find the derivative of the function
To analyze the behavior of \( f(x) \), we differentiate it:
\[
f'(x) = \frac{d}{dx}(\log(1 + x)) - \frac{d}{dx}(x) = \frac{1}{1 + x} - 1
\]
This simplifies to:
\[
f'(x) = \frac{1 - (1 + x)}{1 + x} = \frac{-x}{1 + x}
\]
### Step 3: Determine where the derivative is positive or negative
The sign of \( f'(x) \) tells us about the increasing or decreasing nature of \( f(x) \):
- \( f'(x) > 0 \) when \( x < 0 \) (function is increasing)
- \( f'(x) < 0 \) when \( x > 0 \) (function is decreasing)
### Step 4: Find critical points
Next, we check the value of \( f(x) \) at \( x = 0 \):
\[
f(0) = \log(1 + 0) - 0 = \log(1) = 0
\]
Thus, \( f(0) = 0 \).
### Step 5: Analyze the intervals
Now, we analyze the function in the intervals:
1. For \( x \in (-1, 0) \):
- Since \( f'(x) > 0 \), \( f(x) \) is increasing.
- At \( x = -1 \), \( f(-1) = \log(0) \) which is undefined, but as \( x \) approaches -1 from the right, \( f(x) \) approaches \( -\infty \).
- Therefore, \( f(x) < 0 \) for \( x \in (-1, 0) \).
2. For \( x \in (0, \infty) \):
- Since \( f'(x) < 0 \), \( f(x) \) is decreasing.
- At \( x = 0 \), \( f(0) = 0 \).
- As \( x \) increases, \( f(x) < 0 \) for all \( x > 0 \).
### Step 6: Combine the results
From the analysis:
- \( f(x) \leq 0 \) for \( x \in (-1, 0] \) and for \( x > 0 \).
### Final Answer
Thus, the set of all \( x \) for which \( \log(1 + x) \leq x \) is:
\[
x \in (-1, \infty)
\]
