Q.Find the intervals in which the following functions are strictly increasing or decreasing:
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Increasing Function Test
The Intuition: What Does "Increasing" Really Mean?
Imagine walking along the graph of a function from left to right. If the function is increasing, then as you step right (increasing x), you always move upward — your height f(x) never drops. You might stay flat briefly, but you never go down.
That's the visual idea. But we need a precise way to check it without drawing the entire graph — that's the Increasing Function Test, which uses the derivative to tell you where a function is rising.
A function f is increasing on an interval if, for any two points x1<x2 in it, f(x1)≤f(x2). With strict inequality (<), it's strictly increasing.
The Core Idea: Derivative as a Slope Detector
The derivative f′(x) gives the slope of the tangent line — the instantaneous rate of change. Positive slope means the function is rising at that instant; negative means falling. So the natural question: if the derivative is positive everywhere on an interval, does that guarantee the function is increasing on that whole interval? The answer is yes — and that's the Increasing Function Test.
The Precise Statement
Increasing Function Test
Let f be continuous on [a,b] and differentiable on (a,b).
- If f′(x)>0 for every x in (a,b), then f is strictly increasing on [a,b].
- If f′(x)≥0 for every x in (a,b), then f is increasing (non-decreasing) on [a,b].
The conditions "continuous on the closed interval" and "differentiable on the open interval" ensure there are no jumps or corners that could break the logic.
Why Does This Work? (A Quick Proof Sketch)
The proof relies on the Mean Value Theorem. For x1<x2 in [a,b], there exists some c between them such that:
f(x2)−f(x1)=f′(c)(x2−x1)
Since x2−x1>0, if f′(c)>0 the right-hand side is positive, so f(x2)>f(x1). This holds for any pair x1<x2 — exactly the definition of strictly increasing.
The converse is not true. A function can be strictly increasing even if its derivative is zero at some isolated points. Example: f(x)=x3 is strictly increasing everywhere, but f′(0)=0. The test gives a sufficient condition, not a necessary one.
How to Use It in Practice
- Compute f′(x).
- Solve f′(x)>0 — the solution intervals tell you where f is strictly increasing.
- Check endpoints if needed. …
Concept: Increasing Function Test
A function f is strictly increasing where f′(x)>0 and strictly decreasing where f′(x)<0.
(a) f(x)=x2+2x−5
f′(x)=2x+2=2(x+1). …
For each function, we compute the derivative, find its critical points, and apply the Increasing Function Test (f′(x)>0 for increasing, f′(x)<0 for decreasing) to determine the intervals. The results are given in the final answer block.
The core idea is simple: a function is strictly increasing where its derivative is positive, and strictly decreasing where its derivative is negative. This works because the derivative measures the instantaneous rate of change — if it's positive, the function is climbing; if negative, it's falling. We just need to find where the derivative changes sign, which happens at its zeros (critical points).
Let’s work through each part step by step.
(a) f(x)=x2+2x−5
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Find the derivative:
f′(x)=2x+2=2(x+1)
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Find critical points:
Set f′(x)=0: 2(x+1)=0⇒x=−1. This is the only point where the derivative could change sign.
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Test intervals around x=−1:
- For x<−1, say x=−2: f′(−2)=2(−2+1)=−2<0 → decreasing.
- For x>−1, say x=0: f′(0)=2(0+1)=2>0 → increasing.
-
Conclusion:
Strictly decreasing on (−∞,−1), strictly increasing on (−1,∞).
Don't forget that at x=−1 itself, the derivative is zero — the function is neither strictly increasing nor strictly decreasing at that single point. The intervals are open.
(b) f(x)=10−6x−2x2
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Derivative:
f′(x)=−6−4x=−2(3+2x)
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Critical point:
−2(3+2x)=0⇒x=−23
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Test intervals:
- For x<−23, say x=−2: f′(−2)=−6−4(−2)=−6+8=2>0 → increasing.
- For x>−23, say x=0: f′(0)=−6<0 → decreasing.
-
Conclusion:
Increasing on (−∞,−23), decreasing on (−23,∞).
Notice that this is a downward-opening parabola (−2x2), so it increases up to the vertex and then decreases — exactly what we found.
(c) f(x)=−2x3−9x2−12x+1
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Derivative:
f′(x)=−6x2−18x−12=−6(x2+3x+2)=−6(x+1)(x+2)
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Critical points:
−6(x+1)(x+2)=0⇒x=−1 and x=−2.
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Test intervals: The critical points divide the real line into three intervals: (−∞,−2), (−2,−1), (−1,∞).
- For x<−2, say x=−3: f′(−3)=−6(−3+1)(−3+2)=−6(−2)(−1)=−12<0 → decreasing.
- For −2<x<−1, say x=−1.5: f′(−1.5)=−6(−1.5+1)(−1.5+2)=−6(−0.5)(0.5)=1.5>0 → increasing.
- For x>−1, say x=0: f′(0)=−6(1)(2)=−12<0 → decreasing.
-
Conclusion:
Decreasing on (−∞,−2) and (−1,∞), increasing on (−2,−1).
A cubic can have two turning points — here we see a local minimum at x=−2 and a local maximum at x=−1.
(d) f(x)=6−9x−x2
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Derivative:
f′(x)=−9−2x=−(2x+9)
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Critical point:
−(2x+9)=0⇒x=−29
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Test intervals:
- For x<−29, say x=−5: f′(−5)=−9−2(−5)=−9+10=1>0 → increasing.
- For x>−29, say x=0: f′(0)=−9<0 → decreasing.
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Conclusion:
Increasing on (−∞,−29), decreasing on (−29,∞).
(e) f(x)=(x+1)3(x−3)3
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Simplify first:
f(x)=[(x+1)(x−3)]3=(x2−2x−3)3
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Derivative using chain rule:
f′(x)=3(x2−2x−3)2⋅(2x−2)=3(x2−2x−3)2⋅2(x−1)=6(x−1)(x2−2x−3)2 …
Method: Classifying Increasing/Decreasing Intervals for Any Differentiable Function, Including Repeated Factors
This question covers several different function shapes (upward and downward parabolas, cubics, and a product with repeated linear factors) with one unified method: differentiate, factor completely, and read the sign of each factor — paying special attention to factors that appear to an even power, since those never change sign.
Steps
Step 1: Differentiate each function
Apply the standard rules (power rule, product/chain rule as needed) to get f′(x) for the given expression.
Step 2: Factor f′(x) into simplest linear (or repeated linear) factors
Pull out common constants, then factor any quadratic or higher-degree part fully. Watch for a derivative that comes out as a product involving a squared or otherwise even-power factor, such as (x−a)2 — this happens naturally when the chain rule is applied to something like [g(x)]3.
Step 3: Identify which factors can change sign and which cannot
A factor raised to an odd power (like (x−a) or (x−a)3) changes sign as x crosses a. A factor raised to an even power (like (x−a)2) is always ≥0 and never changes sign — it only touches zero momentarily at x=a without flipping. Only the odd-power factors actually control where the overall sign of f′(x) switches.
Step 4: Build the sign chart using only the sign-changing factors …
Common Mistakes
Mistake 1: Treating a squared factor like (x+1)2(x−3)2 as changing sign at its roots
Why it's wrong: In part (e), f′(x)=6(x−1)(x+1)2(x−3)2 has factors (x+1)2 and (x−3)2 raised to even powers — these are always ≥0 and never actually flip from positive to negative as x crosses −1 or 3, even though f′(x)=0 at those points. Treating them like ordinary sign-changing factors (as with (x−3) in part (c)) leads to a wrong sign chart with extra false intervals. Correct approach: only the odd-power factor (x−1) controls where the overall sign of f′(x) actually switches; the squared factors just make the derivative momentarily touch zero without a genuine sign change.
Mistake 2: Mixing up the sign rule for a downward-opening quadratic derivative, as in parts (b) and (d) …
Showing the 12 most recent of 13 on this concept.
- AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQQ.The set of all real values of 'a' such that the real valued function f(x)=x3+2ax2+3(a+1)x+5 is strictly increasing in its entire domain is (A) (−∞,−43)∪(3,∞) (B) (−43,3) (C) (1,3) (D) (−∞,1)∪(3,∞)
›Reveal solutionSolution
A cubic is strictly increasing on all of R exactly when its derivative (a quadratic with positive leading coefficient) never goes negative, i.e. has non-positive discriminant. This gives a∈(−43,3).
Concept and Intuition
f is strictly increasing everywhere iff f′(x)≥0 for all x (with equality only at isolated points). For a quadratic f′(x)=3x2+4ax+3(a+1) with positive leading coefficient, this non-negativity for all x is equivalent to the discriminant being ≤0 — otherwise the parabola would dip below the x-axis somewhere.
Step-by-Step Solution
- f(x)=x3+2ax2+3(a+1)x+5⇒f′(x)=3x2+4ax+3(a+1).
- Require discriminant of f′ ≤0: (4a)2−4(3)(3(a+1))≤0⇒16a2−36a−36≤0.
- Divide by 4: 4a2−9a−9≤0.
- Solve 4a2−9a−9=0: a=89±81+144=89±15, giving a=3 or a=−43. …
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.Which one of the following functions is monotonically increasing in its domain? (A) f(x)=log(1+x)−x+2x2 (B) g(x)=2Tan−1x−x−1 (C) h(x)=4cosx+x (D) u(x)=log(1+x)−x+1x
›Reveal solutionSolution
A function is monotonically increasing on its domain exactly when its derivative is ≥0 throughout that domain. Testing all four, only option (A)'s derivative stays non-negative everywhere it's defined.
Concept and Intuition
To check monotonicity, differentiate each candidate and determine the sign of the derivative across the entire stated domain — not just at a few sample points. A function can look "mostly increasing" but fail at some sub-interval, which disqualifies it.
Step-by-Step Solution
(A) f(x)=log(1+x)−x+2x2, domain x>−1.
f′(x)=1+x1−1+x=1+x1−(1+x)+x(1+x)=1+x1−1−x+x+x2=1+xx2
Since x>−1⇒1+x>0, and x2≥0 always, f′(x)≥0 throughout the domain (zero only at the single point x=0). So f is monotonically (non-strictly, but genuinely) increasing on its whole domain.
(B) g(x)=2tan−1x−x−1.
g′(x)=1+x22−1=1+x22−(1+x2)=1+x21−x2
This is negative whenever ∣x∣>1, so g decreases for large ∣x∣ — not monotonic on all of R.
(C) h(x)=4cosx+x.
h′(x)=−4sinx+1
Whenever sinx>41 (which happens periodically), h′(x)<0 — not monotonic.
(D) u(x)=log(1+x)−x+1x, domain x>−1. …
- AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQQ.If f(x)=kx3−9x2+9x+3 (k>0) is increasing for all x, then ____ (A) k≤3 (B) k≥3 (C) 0<k<1 (D) 1<k<3
›Reveal solutionSolution
"Increasing for all x" means f′(x)≥0 everywhere; since f′ is an upward parabola in x, this forces its discriminant to be non-positive.
Concept and Intuition
A differentiable function is (weakly) increasing on R iff its derivative is non-negative everywhere. Here f′(x) is itself a quadratic in x; an upward-opening quadratic is non-negative everywhere exactly when it has no two distinct real roots, i.e., discriminant ≤0.
Step-by-Step Solution
- f(x)=kx3−9x2+9x+3⇒f′(x)=3kx2−18x+9.
- Since k>0, f′(x) opens upward. We need f′(x)≥0 ∀x, so discriminant D≤0.
- D=(−18)2−4(3k)(9)=324−108k. …
- AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQQ.If f(x)=xex(1−x), x∈R, then f(x) is (A) increasing on [−21,1] (B) decreasing on R (C) increasing on R (D) decreasing on [−21,1]
›Reveal solutionSolution
The sign of f′(x) is governed by the quadratic factor 1+x−2x2, which is non-negative exactly on [−1/2,1]. Answer: f is increasing on [−21,1].
Concept and Intuition
A function is increasing on an interval where f′≥0. Since f(x)=xex(1−x) is a product of x and an exponential, the product rule brings down an extra polynomial factor from differentiating the exponent, and because e(⋅)>0 always, the entire sign behaviour of f′ reduces to studying that leftover quadratic factor.
Step-by-Step Solution
- Write f(x)=xex−x2 (since x(1−x)=x−x2).
- Product rule: f′(x)=ex−x2+xex−x2(1−2x)=ex−x2[1+x(1−2x)]=ex−x2(1+x−2x2).
- Since ex−x2>0 for all real x, the sign of f′(x) equals the sign of q(x)=1+x−2x2=−(2x2−x−1)=−(2x+1)(x−1).
- 2x2−x−1=0 at x=41±3, i.e. x=1 or x=−21. This upward parabola is ≤0 between its roots, so 2x2−x−1≤0 for x∈[−21,1], hence q(x)=−(2x2−x−1)≥0 there. …
- AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQQ.If g(x)=61f(3x2−1)+21f(1−x2), ∀x∈R, where f′′(x)>0, ∀x∈R. Then g(x) is increasing in the interval ______ (A) (2−1,0)∪(21,∞) (B) (2−1,21) (C) (−1,0)∪(1,2) (D) (−∞,2−1)∪(21,∞)
›Reveal solutionSolution
This uses convexity of f (via f′′>0) to compare f′ at two different points without knowing f explicitly. The answer is (−21,0)∪(21,∞).
Concept and Intuition
Because f′′(x)>0 everywhere, f′ is a strictly increasing function. That means we never need to know f itself — we only need to compare the arguments 3x2−1 and 1−x2 to know which of f′(3x2−1), f′(1−x2) is larger, since a strictly increasing function preserves order.
Step-by-Step Solution
- Differentiate g(x)=61f(3x2−1)+21f(1−x2) using the chain rule:
g′(x)=61f′(3x2−1)(6x)+21f′(1−x2)(−2x)=xf′(3x2−1)−xf′(1−x2)
- So g′(x)=x[f′(3x2−1)−f′(1−x2)].
- Since f′′>0, f′ is strictly increasing, so f′(a)−f′(b) has the same sign as a−b. Here a−b=(3x2−1)−(1−x2)=4x2−2.
- So g′(x) has the same sign as x(4x2−2)=2x(2x2−1), i.e. the same sign as x(2x2−1). …
- AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQQ.The interval in which the curve represented by f(x)=2x+log(2+xx) is increasing is (A) (−∞,0) (B) (−2,∞) (C) (−∞,−2)∪(0,∞) (D) (−2,0)
›Reveal solutionSolution
The function's very domain is (−∞,−2)∪(0,∞), and its derivative simplifies to a manifestly non-negative expression there, so f is increasing on that whole domain.
Concept and Intuition
Before studying monotonicity, always nail down the domain first — a logarithm argument must be strictly positive. Then compute the derivative and factor it; a perfect-square numerator over a product denominator often reveals the sign cleanly without case-by-case sign charts.
Step-by-Step Solution
- Domain: 2+xx>0⟺x and x+2 have the same sign ⟺x>0 or x<−2. So domain =(−∞,−2)∪(0,∞).
- f′(x)=2+x1−x+21=2+x(x+2)(x+2)−x=2+x(x+2)2.
- Combine: f′(x)=x(x+2)2x(x+2)+2=x(x+2)2(x2+2x+1)=x(x+2)2(x+1)2.
- On x>0: x(x+2)>0, and (x+1)2≥0, so f′(x)≥0.
- On x<−2: both x<0 and x+2<0, so x(x+2)>0 again, and f′(x)≥0. …
- AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQQ.The interval containing all the real values of x such that the real valued function f(x)=x+x1 is strictly increasing is (A) (1,∞) (B) (0,1) (C) (−∞,0)∪(1,∞) (D) (−∞,0)
›Reveal solutionSolution
Differentiate and check the sign on the natural domain x>0; f strictly increases on (1,∞).
Concept and Intuition
A function is strictly increasing wherever its derivative is (strictly) positive. Since x1 requires x>0, the domain of f is restricted to positive reals from the start — there's no negative-x branch to worry about.
Step-by-Step Solution
- Domain: x needs x≥0 and x1 needs x>0, so the domain is (0,∞).
- f′(x)=2x1−21x−3/2=2x1(1−x1)=2x1⋅xx−1=2x3/2x−1.
- On (0,∞), 2x3/2>0 always, so the sign of f′(x) matches the sign of (x−1). …
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.The function f(x)=x2+x54 (A) is increasing and has minimum value 27 in the interval (0,∞) (B) is decreasing and has neither maximum nor minimum in the interval (−∞,0) (C) has maximum value 27 in the interval (−∞,∞) (D) is increasing and has neither maximum nor minimum values in the interval (−∞,∞)
›Reveal solutionSolution
Sign-analysis of f′(x)=2x−54/x2 shows f is strictly decreasing throughout (−∞,0) with no turning point there (the only critical point x=3 lies in (0,∞)), so option (B) is the true statement.
Concept and Intuition
A function is monotonic on an interval exactly when its derivative keeps one sign throughout that interval; local extrema only occur where the derivative is zero (or undefined) and changes sign. Checking (−∞,0) and (0,∞) separately (since f isn't even defined at x=0) settles all four options at once.
Step-by-Step Solution
- f(x)=x2+x54⇒f′(x)=2x−x254.
- Critical points: f′(x)=0⇒2x=x254⇒2x3=54⇒x3=27⇒x=3 (the only real root).
- On (0,∞): for 0<x<3, e.g. x=1: f′(1)=2−54=−52<0; for x>3, e.g. x=4: f′(4)=8−54/16>0. So f decreases on (0,3) then increases on (3,∞) — a genuine local minimum at x=3, value f(3)=9+18=27, but f is not monotonically increasing throughout (0,∞) — rules out (A).
- On (−∞,0): for any x<0, 2x<0 and −54/x2<0 (since x2>0 always makes −54/x2 negative) — so f′(x)<0 for every x<0. Thus f is strictly decreasing on all of (−∞,0), with no sign change, hence no interior local max or min. …
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.If the function f(x)=sinx−cos2x is defined on the interval [−π,π], then f is strictly increasing in the interval (A) (6−5π,6−π)∪(6−π,2π) (B) (2−π,6−π) (C) (6−5π,2π) (D) (6−5π,2−π)∪(6−π,2π)
›Reveal solutionSolution
Factoring f′(x)=cosx(1+2sinx) and doing a careful sign analysis over [−π,π] shows f is strictly increasing on (−65π,−2π)∪(−6π,2π).
Concept and Intuition
A function is strictly increasing exactly where its derivative is strictly positive. Since f′ here factors into two simple trig expressions, the problem reduces to tracking the signs of cosx and (1+2sinx) separately across the interval and multiplying the signs region by region.
Step-by-Step Solution
- f(x)=sinx−cos2x⇒f′(x)=cosx−2cosx(−sinx)=cosx+2sinxcosx=cosx(1+2sinx).
- Find zeros of each factor in [−π,π]:
- cosx=0 at x=−2π,2π.
- 1+2sinx=0⇒sinx=−21 at x=−65π,−6π.
- These four points split [−π,π] into five intervals: (−π,−65π), (−65π,−2π), (−2π,−6π), (−6π,2π), (2π,π).
- Test the sign of f′(x)=cosx(1+2sinx) in each:
- (−π,−65π): cosx<0, sinx near 0 so 1+2sinx>0 ⇒f′<0.
- (−65π,−2π): cosx<0, sinx<−21 so 1+2sinx<0 ⇒f′>0.
- (−2π,−6π): cosx>0, sinx<−21 so 1+2sinx<0 ⇒f′<0.
- (−6π,2π): cosx>0, sinx>−21 so 1+2sinx>0 ⇒f′>0. …
- AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQQ.y=x3−ax2+48x+7 is an increasing function for all real values of x, then a lies in the interval (A) (−14,14) (B) (−12,12) (C) (−16,16) (D) (−21,−21)
›Reveal solutionSolution
A cubic is increasing everywhere exactly when its derivative (an upward parabola) never goes negative — that discriminant condition gives a∈(−12,12).
Concept and Intuition
y=x3−ax2+48x+7 increases everywhere iff y′(x)≥0 for every real x. Since y′=3x2−2ax+48 is an upward-opening parabola (positive leading coefficient), it stays ≥0 for all x exactly when its discriminant is ≤0 (no real roots, or a repeated root, so it never dips below zero).
Step-by-Step Solution
- y′=3x2−2ax+48.
- Require y′≥0 ∀x: discriminant ≤0: (−2a)2−4(3)(48)≤0.
- 4a2−576≤0⇒a2≤144⇒−12≤a≤12.
Common Mistakes …
- AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQQ.In the interval (7,∞), f(x)=∣x−5∣+2∣x−7∣ is (A) increasing function (B) decreasing function (C) constant function (D) attains maximum value
›Reveal solutionSolution
On the interval (7,∞) both absolute-value expressions can be opened without a sign flip, reducing f(x) to the simple linear function 3x−19, which is clearly increasing.
Concept and Intuition
Absolute value functions are piecewise linear, with "kinks" only at the points where the inner expression changes sign (here at x=5 and x=7). To analyze behaviour on (7,∞) we just need to know the sign of each inner expression throughout that interval — beyond both kink points, both expressions are positive, so the absolute values open up with a plain + sign.
Step-by-Step Solution
- For x>7: since x>7>5, we have x−5>0 and x−7>0.
- So ∣x−5∣=x−5 and ∣x−7∣=x−7 on this interval.
- f(x)=(x−5)+2(x−7)=x−5+2x−14=3x−19.
- f′(x)=3>0 for all x in (7,∞), so f is strictly increasing there.
Common Mistakes …
- AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQQ.If f′′(x) is a positive function for all x∈R, f′(3)=0 and g(x)=f(tan2(x)−2tan(x)+4) for 0<x<2π, then the interval in which g(x) is increasing is ______. (A) (6π,3π) (B) (0,4π) (C) (0,3π) (D) (4π,2π)
›Reveal solutionSolution
Complete the square in u to see u≥3 always (so f′(u)≥0), then the sign of g′ is controlled entirely by u′(x), which turns positive past x=π/4. Answer: (4π,2π).
Concept and Intuition
Since f′′>0, f′ is strictly increasing, and f′(3)=0 means f′(t)>0 for t>3 and f′(t)<0 for t<3. If we can show the inner function u(x) never goes below 3, then f′(u)≥0 everywhere and the composite's monotonicity is governed purely by u′(x)'s sign.
Step-by-Step Solution
- u(x)=tan2x−2tanx+4=(tanx−1)2+3, which is always ≥3, with equality iff tanx=1 i.e. x=π/4.
- So f′(u(x))≥0 for all x∈(0,π/2), with f′(u)=0 only exactly at x=π/4.
- g′(x)=f′(u(x))⋅u′(x), where u′(x)=2tanxsec2x−2sec2x=2sec2x(tanx−1).
- sec2x>0 always, so sign(u′(x))=sign(tanx−1): negative for x<π/4, positive for x>π/4. …
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