15. The function x + sin(x) is best described as
a. non-decreasing
b. non-increasing
c. decreasing
d. increasing
Is x + sin(x) Non-Decreasing? Monotonicity Analysis & Derivative Proof
The function f(x)=x+sin x is non-decreasing because its derivative f′(x)=1+cos x≥0 for all real x, with equality only at isolated points where cos x=−1. This makes option a. non-decreasing correct. The correct answer is a.
Option Analysis
Non-decreasing: Holds as f′(x)≥0 everywhere, so f(x1)≤f(x2) for x1≤x2.
Non-increasing: Incorrect, since f′(x)>0 over most intervals, causing the function to rise overall.
Decreasing: Fails because f′(x) never stays negative; the slope is always non-negative.
Increasing: Technically strict increasing requires f′(x)>0 everywhere, but zeros at points like x=(2k+1)π make it non-strictly increasing, fitting non-decreasing best.
The x + sin(x) function exhibits fascinating monotonicity properties central to calculus exams. Often queried as “is x + sin(x) non-decreasing?”, this analysis reveals its behavior through derivatives, perfect for JEE/competitive math prep.
Derivative Proof
Compute f(x)=x+sin x, so f′(x)=1+cos x. Since −1≤cos x≤1, then 0≤f′(x)≤2. The derivative equals zero only at isolated points x=(2k+1)π (where cos x=−1), but remains positive elsewhere, confirming non-decreasing nature across R.
Monotonicity Breakdown
| Property | Condition | Applies to x + sin(x)? | Reason |
|---|---|---|---|
| Strictly Increasing | f′(x)>0 everywhere | No | Zeros in derivative |
| Non-Decreasing | f′(x)≥0 everywhere | Yes | 1+cos x≥0 |
| Strictly Decreasing | f′(x)<0 everywhere | No | Derivative never negative |
| Non-Increasing | f′(x)≤0 everywhere | No | Often positive slope |
Applications in Exams
In MCQs like “x + sin(x) is best described as”, select non-decreasing over increasing due to precise definitions. Visualize: the linear x dominates oscillations of sin x.
