Q.Show that the function f(x)=log(1+x)+1+x1 increases on (0,∞).
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.
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 x1<x2 in I, we have f(x1)≤f(x2).
- f is strictly increasing on I if, for any x1<x2 in I, we have f(x1)<f(x2).
- f is decreasing (or non-increasing) on I if, for any x1<x2 in I, we have f(x1)≥f(x2).
- f is strictly decreasing on I if, for any x1<x2 in I, we have f(x1)>f(x2).
If a function is either increasing or decreasing on an interval, it is called monotonic on that interval.
| 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)=x3 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)=x2 is decreasing on (−∞,0] and increasing on [0,∞), but not monotonic on R. |
How to Determine Monotonicity (The Derivative Test)
For differentiable functions, the derivative tells you the direction:
- If f′(x)≥0 for all x in an interval, then f is increasing on that interval.
- If f′(x)≤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)≥0⟹f increasingf′(x)≤0⟹f 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 ex>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)=x3−3x+1.
Solution:
First, find the derivative:
f′(x)=3x2−3=3(x2−1)=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 |
|---|---|---|---|
| (−∞,−1) | x=−2 | 3(4−1)=9>0 | Increasing |
| (−1,1) | x=0 | 3(0−1)=−3<0 | Decreasing |
| (1,∞) | x=2 | 3(4−1)=9>0 | Increasing |
Therefore, f is increasing on (−∞,−1] and [1,∞), and decreasing on [−1,1].
| 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.
To show f is increasing on (0,∞), it suffices to compute f′(x) and confirm it stays positive throughout that interval.
f′(x)=(1+x)2x>0 for all x>0, so f is increasing on (0,∞).
f′(x) simplifies to (1+x)2x, which is positive for every x>0; hence f increases on (0,∞).
f is increasing on an interval if f′(x)>0 throughout it.
- Given f(x)=log(1+x)+1+x1.
- Differentiate: dxdlog(1+x)=1+x1 and dxd(1+x)−1=−(1+x)21.
- So f′(x)=1+x1−(1+x)21.
- Take LCD (1+x)2: f′(x)=(1+x)2(1+x)−1=(1+x)2x.
- For x>0: numerator x>0 and denominator (1+x)2>0, so f′(x)>0.
- Since f′(x)>0 for all x∈(0,∞), f is (strictly) increasing on (0,∞).
f′(x)=(1+x)2x>0 for all x>0; therefore f increases on (0,∞). Proved.
- CBSE 2025Set 465/S/WXYZ/41 markMCQQ.The function f(x)=ax is increasing on R, if : (A) a>0 (B) a>1 (C) a<0 (D) 0<a<1
›Reveal solutionSolution
ax is increasing on R iff a>1, since that makes lna>0 and hence f′(x)>0 everywhere.
dxd(ax)=axlna; a function is increasing where its derivative is positive.
- Differentiate: f′(x)=axlna.
- For all real x, ax>0 (with a>0), so the sign of f′(x) is the sign of lna.
- f′(x)>0 for all x⟺lna>0⟺a>1.
- (If 0<a<1, lna<0 and f decreases; at a=1 it is constant.)
✓Final answer(B) a>1
- CBSE 2024Set 465/RQPS/41 markMCQQ.The function f(x)=x2−x+1 is : (A) increasing in (0,1) (B) decreasing in (0,1) (C) increasing in (0,21) and decreasing in (21,1) (D) increasing in (21,1) and decreasing in (0,21)
›Reveal solutionSolution
The turning point is at x=21; f decreases on (0,21) and increases on (21,1).
f is increasing where f′(x)>0 and decreasing where f′(x)<0.
- f(x)=x2−x+1⇒f′(x)=2x−1.
- f′(x)=0⇒x=21 (the critical point).
- For x∈(0,21), f′(x)<0 — f is decreasing.
- For x∈(21,1), f′(x)>0 — f is increasing.
✓Final answer(D) increasing in (21,1) and decreasing in (0,21)
- CBSE 2023Set 465/EF1GH/41 markMCQQ.Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b),(c) and(d) as given below.(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(c) Assertion (A) is true and Reason (R) is false.(d) Assertion (A) is false and Reason (R) is true. Assertion (A) : The function f(x)=(x+2)e−x is increasing in the interval (−1,∞). Reason (R) : A function f(x) is increasing, if f′(x)>0.
›Reveal solutionSolution
f′(x)=−(x+1)e−x>0 only for x<−1, so f is decreasing on (−1,∞) — A is false; R is a correct general statement — so answer (d).
A differentiable function is increasing on an interval where f′(x)>0. Product rule: (uv)′=u′v+uv′.
- f(x)=(x+2)e−x. Differentiate: f′(x)=(1)e−x+(x+2)(−e−x)=e−x[1−(x+2)]=−(x+1)e−x.
- Since e−x>0 always, sign(f′)=sign(−(x+1)).
- f′(x)>0⟺x+1<0⟺x<−1. So f increases on (−∞,−1) and decreases on (−1,∞).
- Therefore Assertion (claiming increasing on (−1,∞)) is FALSE; Reason ('f increasing if f′(x)>0') is TRUE. A false, R true ⇒ code (d).
✓Final answer(d) Assertion (A) is false and Reason (R) is true
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.