Q.Prove that the function given by f(x)=x3−3x2+3x−5 is increasing in R.
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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. …
A function is increasing wherever its derivative is non-negative, so showing f'(x) can be written as a perfect square (which is never negative) proves the function is increasing across the whole real li …
show the derivative is a non-negative perfect square everywhere
f(x)=x3−3x2+3x−5⟹f′(x)=3x2−6x+3=3(x2−2x+1)=3(x−1)2
Since (x−1)2≥0 for every real x, f′(x)≥0 for all x∈R (equality only at the isolated point x=1).
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Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set CX1 markMCQQ.Interval in which the given function f(x)=x2−4x+6 is increasing, is:(a) (2,10)(b) (2,∞)(c) (−2,∞)(d) (0,∞)
›Reveal solutionSolution
f′(x)=2x−4 is positive for x>2, so f is increasing on (2,∞) — option (b).
Concept: A differentiable function increases where its derivative is positive.
f(x)=x2−4x+6⇒f′(x)=2x−4.
Set f′(x)>0:
2x−4>0⇒x>2. …
- CBSE 2026Set ANNUAL1 markQ.Show that the function f(x) = x³ − 3x² + 3x + 10 is always increasing.
›Reveal solutionSolution
A function is increasing on an interval where its derivative is non-negative there; we show f′(x)≥0 for every real x.
f(x)=x3−3x2+3x+10
f′(x)=3x2−6x+3=3(x2−2x+1)=3(x−1)2
…
- CBSE 2026Set ANNUAL1 markMCQQ.Assertion (A): f(x)=x4 is decreasing in the interval (0,∞). Reason (R): Any derivable function y=f(x) is decreasing if dxdy<0. Answer by selecting the appropriate option:(a) Both A and R are true and R is the correct explanation of A(b) Both A and R are true and R is not the correct explanation of A(c) A is true but R is false(d) A is false but R is true
›Reveal solutionSolution
f(x)=x4 has f′(x)=4x3>0 on (0,∞), so it is increasing (A false). The Reason (negative derivative ⇒ decreasing) is a true criterion (R true).
Assertion: f′(x)=4x3. For x∈(0,∞), f′(x)>0, so f is increasing, not decreasing. Hence A is false.
…
- CBSE 2025Set 65/2/11 markMCQQ.The function f(x)=x2−4x+6 is increasing in the interval: (A) (0,2) (B) (−∞,2] (C) [1,2] (D) [2,∞)
›Reveal solutionSolution
The function f(x)=x2−4x+6 is a parabola opening upward, so it decreases until its vertex and then increases. The vertex is at x=2, so the function is increasing on [2,∞). The correct option is (D).
The key idea here is the Increasing Function Test from calculus: a function f(x) is increasing on an interval if its derivative f′(x)≥0 for all x in that interval (and strictly increasing if f′(x)>0). But before we dive into derivatives, let's think about what this function looks like.
f(x)=x2−4x+6 is a quadratic — a parabola. The coefficient of x2 is positive (it's 1), so the parabola opens upward. That means it has a single minimum point (the vertex), falls to the left of that vertex, and rises to the right. So the function is decreasing on (−∞,vertex] and increasing on [vertex,∞). The question is simply: where is the vertex?
Let's work through it step by step.
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Find the derivative.
f′(x)=2x−4. This is a linear function. The sign of f′(x) tells us where f is increasing or decreasing.
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Set the derivative to zero to find the critical point.
2x−4=0⟹x=2. This is the vertex of the parabola — the point where the function stops decreasing and starts increasing.
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Test the sign of f′(x) on either side of x=2.
- For x<2, say x=0: f′(0)=−4<0. So f is decreasing on (−∞,2).
- For x>2, say x=3: f′(3)=2>0. So f is increasing on (2,∞).
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What about at x=2 itself?
f′(2)=0. The function is neither increasing nor decreasing at that single point, but by convention, we include the endpoint where the derivative is zero when describing intervals of monotonicity. So the function is increasing on [2,∞). …
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- CBSE 2025Set 65/4/11 markMCQQ.If f(x)=2x+cosx, then f(x) : (A) has a maxima at x=π (B) has a minima at x=π (C) is an increasing function (D) is a decreasing function
›Reveal solutionSolution
A function is increasing when its derivative is always positive. Since f′(x)=2−sinx≥1>0 for all x, the function is strictly increasing everywhere.
The question asks about the monotonicity and extrema of f(x)=2x+cosx. To understand the behavior of any function, we look at its derivative: the sign of f′(x) tells us whether the function is climbing or falling at each point.
A function has a local maximum or minimum only where f′(x)=0 (critical points), and even then only if the derivative changes sign. If f′(x) never changes sign—if it's always positive or always negative—the function marches steadily in one direction without any peaks or valleys.
Let me find the derivative and analyze its sign.
- Differentiate f(x):
f′(x)=dxd(2x+cosx)=2−sinx
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Examine the range of f′(x):
We know that sinx oscillates between −1 and 1 for all real x. Therefore:
−1≤sinx≤1
Multiplying by −1 (which reverses inequalities):
−1≤−sinx≤1
Adding 2 throughout:
1≤2−sinx≤3
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Interpret the result:
The derivative f′(x)=2−sinx satisfies 1≤f′(x)≤3 for all x. In particular, f′(x)≥1>0 everywhere.
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Conclude about monotonicity: …
- CBSE 2025Set A1 markQ.Find the intervals in which the function f given by f(x)=x2−2x is increasing.
›Reveal solutionSolution
Find f′(x) and determine where it is positive.
Given f(x)=x2−2x:
f′(x)=2x−2=2(x−1)
f is increasing where f′(x)>0:
2(x−1)>0⟹x>1
…
- CBSE 2025Set ANNUAL1 markMCQQ.In which interval is the function y=lnx, x∈R+ increasing?(i) (0,∞)(ii) (−∞,∞)(iii) (−∞,0)(iv) (−1,∞)
›Reveal solutionSolution
y=lnx has y′=1/x>0 on its entire domain, so it is increasing on (0,∞).
y=lnx⟹dxdy=x1
For x∈R+ (i.e. x>0), x1>0 always. A function whose derivative is positive throughout an interval is strictly increasing on that interval.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The function f(x) = log x is :(a) Strictly increasing on (0, ∞)(b) Strictly decreasing on (0, ∞)(c) Neither increasing nor decreasing on (0, ∞)(d) None of these
›Reveal solutionSolution
f′(x)=1/x>0 for every x>0, so logx is strictly increasing throughout (0,∞).
A function f is strictly increasing on an interval if f′(x)>0 for all x in that interval. For f(x)=logx (domain x>0),
f′(x)=x1. …
- CBSE 2025Set ANNUAL1 markQ.Show that the function f(x)=3x+17 is strictly increasing on R.
›Reveal solutionSolution
Show the derivative is positive throughout R.
A differentiable function is strictly increasing on an interval where f′(x)>0.
For f(x)=3x+17,
f′(x)=dxd(3x+17)=3.
This is positive for every real x:
f′(x)=3>0∀x∈R. …
- CBSE 2025Set ANNUAL1 markMCQQ.The interval in which f(x)=x2e−x is increasing in(a) (−∞,∞)(b) (−2,0)(c) (2,∞)(d) (0,2)
›Reveal solutionSolution
Differentiate, factor f′(x), and find where it is positive.
Given f(x)=x2e−x.
f′(x)=2xe−x+x2(−e−x)=e−x(2x−x2)=e−xx(2−x)
Since e−x>0 always, the sign of f′(x) is the sign of x(2−x).
x(2−x)>0 when both factors have the same sign:
- x>0 and 2−x>0⇒0<x<2 …
- CBSE 2025Set ANNUAL1 markQ.Find the interval in which the function f(x) = 2x² + 12x + 1 is increasing.
›Reveal solutionSolution
A function is increasing where its first derivative is positive. Find f′(x), set it >0, and solve for x.
Given: f(x)=2x2+12x+1
Step 1 — differentiate:
f′(x)=4x+12
Step 2 — set f′(x)>0 for increasing:
4x+12>0⇒4x>−12⇒x>−3
…
- CBSE 2024Set ANNUAL1 markMCQQ.In which of the following intervals is y=x2e−x increasing?(a) (1,0)(b) (2,0)(c) (2,−∞)(d) (0,2)
›Reveal solutionSolution
Find y′ by the product rule and determine where it is positive.
y=x2e−x
y′=2xe−x−x2e−x=xe−x(2−x)
Since e−x>0 for all x, the sign of y′ matches the sign of x(2−x).
…
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