Monotonic Functions Increasing, Decreasing, Concavity and Inflection

The nature of a function $f(x)$ which is differentiable everywhere can only be one of three things:

• the output increases as the input increases,
• the output decreases as the input increases,
• the output does neither, because the function rises on some stretches and falls on others.

A function is called monotonic when it does the same thing throughout the interval you are looking at. If it rises on one stretch and falls on another, it is monotonic on each stretch separately, but not on the whole. Everything in this chapter is about deciding which stretch is which, and the derivative is the tool that decides it quickly.

Fundamental concept of monotonic functions

Let $f$ be defined on an interval $I$, and take any $x_1,x_2\in I$ with $x_1<x_2$.

Strictly increasing and strictly decreasing functions — the definition on a graph

Geometrical meaning — tangent makes an acute angle when increasing, obtuse when decreasing

Non-decreasing and non-increasing

A function is called non-decreasing (or increasing in the weak sense) if $x_1<x_2\Rightarrow f(x_1)\le f(x_2)$. For two different inputs the function may return the same output, but it never returns a smaller one. Similarly $f$ is non-increasing if $x_1<x_2\Rightarrow f(x_1)\ge f(x_2)$: it may repeat a value, but never grow.

The visible difference is that a non-decreasing function is allowed to stop and sit still on a stretch, while a strictly increasing one is never allowed to.

Non-decreasing and non-increasing functions — flat stretches allowed

Reading monotonicity off the derivative

The two facts above have a converse, and the converse is the one you will actually use. It comes from Lagrange's mean value theorem.

For $x_1<x_2$ in the interval, LMVT supplies a point $c$ between them with

$f(x_2)-f(x_1)=f'(c)\,(x_2-x_1).$

The bracket $x_2-x_1$ is positive, so the sign of $f(x_2)-f(x_1)$ is exactly the sign of $f'(c)$. Hence, for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$:

• $f'(x)>0$ on $(a,b)$ $\Rightarrow$ $f$ is strictly increasing on $[a,b]$,
• $f'(x)<0$ on $(a,b)$ $\Rightarrow$ $f$ is strictly decreasing on $[a,b]$,
• $f'(x)=0$ on $(a,b)$ $\Rightarrow$ $f$ is constant on $[a,b]$.

Sign of the derivative decides monotonicity — f'>0 rising, f'<0 falling

Note the hypotheses: continuity on the closed interval, differentiability only on the open one. That is why $f(x)=\sqrt x$ counts as strictly increasing on $[0,\infty)$ even though $f'(0)$ does not exist.

The one statement that is false

Row 3 of the table gave $f'\ge0$, not $f'>0$. So while $f'>0$ guarantees strictly increasing, strictly increasing does not guarantee $f'>0$. This is the most heavily examined misconception in the chapter.

$f(x)=x^3$ is strictly increasing on $\mathbb R$, which you can prove with no calculus at all: for $x_1<x_2$,

$x_2^3-x_1^3=(x_2-x_1)\left[\left(x_2+\tfrac{x_1}{2}\right)^2+\tfrac{3x_1^2}{4}\right]>0.$

And yet $f'(0)=0$.

y = x cubed is strictly increasing on R yet has f'(0)=0

The correct statement allows the derivative to vanish, provided it does not stay vanished:

If $f'\ge0$ on $I$ and $f'$ does not vanish identically on any sub-interval — in particular if its zeros are isolated — then $f$ is strictly increasing on $I$.

Why: $f'\ge0$ already makes $f$ non-decreasing. If two points had equal images, $f$ would be constant between them, forcing $f'=0$ on that whole sub-interval, which the hypothesis forbids.

Double derivative of monotonic functions

The sign of $f'$ says which way the curve goes. The sign of $f''$ says how it bends, and the two are independent — so an increasing function has three possible shapes, and so does a decreasing one.

$f$ is concave up where $f'$ is increasing, i.e. where $f''>0$; concave down where $f'$ is decreasing, i.e. where $f''<0$. This is just the derivative test applied to $f'$ instead of $f$.

A. Increasing

An increasing curve can bend up ($f''>0$), bend down ($f''<0$), or not bend at all ($f''=0$, a straight line). All three are rising.

Increasing functions — the three shapes from the second derivative

B. Decreasing

The same three shapes occur while falling: bending down ($f''<0$), bending up ($f''>0$), or a straight line ($f''=0$).

Decreasing functions — the three shapes from the second derivative

Point of inflection

For a continuous $f$, a point $x=x_0$ is a point of inflection if the concavity changes as $x$ passes through it — concave up on one side, concave down on the other. If $f''$ is continuous near $x_0$, then an inflection forces $f''(x_0)=0$; conversely, if $f''$ changes sign at $x_0$, then $x_0$ is an inflection point.

Point of inflection — concavity changes where f''(x0)=0

Now the six question types that use all of this.

Type 1 — Monotonicity at a point

Monotonicity at a single point compares $f(a)$ only with its immediate neighbours. $f$ is increasing at $x=a$ if there is a small $h>0$ with

$f(a-h)<f(a)<f(a+h),$

and decreasing at $a$ if $f(a-h)>f(a)>f(a+h)$.

Continuity and differentiability are simply not part of this definition, so a corner or a jump at $x=a$ does not by itself stop $f$ from being increasing at $a$. We may speak of monotonicity of $f$ at $x=a$ only if $a$ lies in the domain of $f$.

Example 1. At $x=a$ the graph of $f$ satisfies $f(a-h)<f(a)$ and $f(a+h)=f(a)$ for all small $h>0$. Classify $f$ at $x=a$.

▸ SHOW SOLUTION

Step 1: Test the increasing chain. It needs $f(a)<f(a+h)$. Here $f(a+h)=f(a)$, so the second inequality is not strict and the chain fails.

Step 2: Test the decreasing chain. It needs $f(a-h)>f(a)$. Here $f(a-h)<f(a)$, so it fails at the first step.

Step 3: Test the weak version. $f(a-h)\le f(a)\le f(a+h)$ does hold.

Answer: $f$ is neither increasing nor decreasing at $a$; it is non-decreasing at $a$.

Type 2 — Monotonicity in an interval

This is the standard case. The method is fixed:

1. Write the domain first — everything that follows is trapped inside it.
2. Differentiate and factorise $f'$.
3. Mark the partition points: zeros of $f'$, and domain points where $f'$ fails to exist.
4. Build a sign chart on the gaps between consecutive partition points.
5. Report the intervals one at a time.

Example 2. Find the intervals in which $f(x)=x^3-6x^2+9x+2$ increases and decreases.

▸ SHOW SOLUTION

Step 1: Domain. $\mathbb R$, since $f$ is a polynomial.

Step 2: Factorise the derivative.

$f'(x)=3x^2-12x+9=3(x-1)(x-3).$

Step 3: Partition points. $x=1$ and $x=3$.

Step 4: Sign chart. The factor $3$ is positive, so the sign of $f'$ is the sign of $(x-1)(x-3)$ — a product of two brackets, positive when both share a sign.

Step 5: Report. Strictly increasing on $(-\infty,1]$ and on $[3,\infty)$; strictly decreasing on $[1,3]$. Endpoints attach because $f$ is continuous and $f'$ vanishes only at those single points.

Sign chart of f'(x) = 3(x-1)(x-3)

Example 3. For which real $a$ is $f(x)=x^3+ax^2+3x+7$ strictly increasing on $\mathbb R$?

▸ SHOW SOLUTION

Step 1: Differentiate. $f'(x)=3x^2+2ax+3$.

Step 2: Decide what to demand. It is enough that $f'\ge0$ everywhere, provided $f'$ does not vanish on a whole interval. A non-zero quadratic has at most two roots, so that is automatic.

Step 3: Discriminant. The leading coefficient $3$ is positive, so $3x^2+2ax+3\ge0$ for all $x$ exactly when

$(2a)^2-4\cdot3\cdot3\le0\ \Longrightarrow\ 4a^2\le36\ \Longrightarrow\ a^2\le9.$

Step 4: Solve. $-3\le a\le3$.

Step 5: Check the endpoints. At $a=3$, $f'(x)=3(x+1)^2\ge0$, vanishing only at $x=-1$ — allowed. Same at $a=-3$.

Answer: $\boxed{a\in[-3,3]}$, a closed interval.

Example 4 (JEE Main 2026, 24 Jan Shift 1). Let $(2\alpha,\alpha)$ be the largest interval in which $f(t)=\dfrac{|t+1|}{t^2}$, $t<0$, is strictly decreasing. Find the local maximum value of $g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha$, $x>2$.

▸ SHOW SOLUTION

Step 1: Locate the switch of the modulus. On $t<0$, the expression $t+1$ is negative for $t<-1$. Since $(2\alpha,\alpha)$ needs $2\alpha<\alpha$, we get $\alpha<0$, so work on $t<-1$ where $|t+1|=-(t+1)$.

Step 2: Differentiate. With $f(t)=-\dfrac1t-\dfrac1{t^2}$,

$f'(t)=\frac{1}{t^2}+\frac{2}{t^3}=\frac{t+2}{t^3}.$

Step 3: Sign chart on $t<-1$. The denominator $t^3<0$ there. The numerator $t+2$ is positive on $(-2,-1)$ and negative on $(-\infty,-2)$. So $f'<0$ on $(-2,-1)$ and $f'>0$ on $(-\infty,-2)$: the largest interval of strict decrease is $(-2,-1)$.

Step 4: Match. $\alpha=-1$, and then $2\alpha=-2$ agrees.

Step 5: Write $g$ and differentiate.

$g(x)=2\log_e(x-2)-x^2+4x+1,\qquad g'(x)=\frac{2}{x-2}-2(x-2).$

Step 6: Critical points with $x>2$. $g'=0$ gives $(x-2)^2=1$, so $x=3$ (the root $x=1$ fails $x>2$).

Step 7: Confirm it is a maximum.

$g'(x)=\frac{2\big(1-(x-2)^2\big)}{x-2},$

whose sign for $x>2$ follows $1-(x-2)^2$: positive on $(2,3)$, negative on $(3,\infty)$. The sign goes $+\to-$, so $x=3$ is a local maximum.

Step 8: Evaluate.

$g(3)=2\log_e1-9+12+1=\boxed{4}$

y = sin 3x — longest rising run has length pi/3

Type 3 — Checking the one-one (injective) nature

If $f$ is strictly increasing or strictly decreasing on a set, it is one-one there: for two distinct inputs, one is smaller, so their images are strictly ordered and cannot be equal.

The converse fails. A function equal to $x$ on $[0,1)$ and to $3-x$ on $[1,2]$ is one-one but neither increasing nor decreasing. So use monotonicity to prove one-one, never to disprove it.

Example 5 (JEE Main 2026, 28 Jan Shift 2). Show that $f:\mathbb R\to\mathbb R$, $f(x)=\dfrac{x}{1+|x|}$, is one-one.

▸ SHOW SOLUTION

Step 1: Remove the modulus.

$f(x)=\begin{cases}\dfrac{x}{1+x}, & x\ge0,\\[3mm] \dfrac{x}{1-x}, & x<0.\end{cases}$

Step 2: Differentiate on $x>0$.

$f'(x)=\frac{(1+x)\cdot1-x\cdot1}{(1+x)^2}=\frac{1}{(1+x)^2}>0.$

Step 3: Differentiate on $x<0$.

$f'(x)=\frac{(1-x)\cdot1-x\cdot(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}>0.$

Step 4: Join the pieces. Both formulas give $f(0)=0$, so $f$ is continuous at $0$. With positive derivative on both sides and only the single point $x=0$ left over, $f$ is strictly increasing on all of $\mathbb R$.

Step 5: Conclude. Strictly increasing $\Rightarrow$ one-one.

A strictly monotonic $f$ is also invertible onto its range, and $f^{-1}$ keeps the same direction: if $y_1<y_2$ but $f^{-1}(y_1)\ge f^{-1}(y_2)$, applying the increasing $f$ would give $y_1\ge y_2$, a contradiction.

Type 4 — Monotonicity of a composite function

Direction composes like multiplication of signs:

Why the second row: for $x_1<x_2$, a decreasing $g$ gives $g(x_1)>g(x_2)$; feeding those into an increasing $f$ preserves the order, so $f(g(x_1))>f(g(x_2))$.

Example 6 (JEE Main 2026, 21 Jan Shift 2). Let $f:\mathbb R\to\mathbb R$ be twice differentiable with $f''(x)>0$ for all $x$ and $f'(a-1)=0$. Put $g(x)=f\left(\tan^2x-2\tan x+a\right)$ on $0<x<\dfrac\pi2$. Decide: (I) $g$ is increasing on $\left(0,\dfrac\pi4\right)$; (II) $g$ is decreasing on $\left(\dfrac\pi4,\dfrac\pi2\right)$.

▸ SHOW SOLUTION

Step 1: Complete the square inside $f$.

$\tan^2x-2\tan x+a=(\tan x-1)^2+(a-1),$

so $g(x)=f\big(u(x)\big)$ with $u(x)=(\tan x-1)^2+(a-1)$.

Step 2: Chain rule. $u'(x)=2(\tan x-1)\sec^2x$, so

$g'(x)=f'\big(u(x)\big)\cdot 2(\tan x-1)\sec^2x.$

Step 3: The factor $f'(u)$ is never negative. Since $f''>0$ everywhere, $f'$ is strictly increasing. And $u(x)\ge a-1$, because a square is $\ge0$. Feeding the larger input into the increasing $f'$:

$f'\big(u(x)\big)\ \ge\ f'(a-1)=0.$

Step 4: Reduce the sign to one bracket. $2\sec^2x>0$ on $\left(0,\dfrac\pi2\right)$ and $f'(u)\ge0$, so

$\operatorname{sign}\big(g'(x)\big)=\operatorname{sign}(\tan x-1).$

Step 5: Evaluate on each interval. On $\left(0,\dfrac\pi4\right)$, $\tan x<1$, so $g'\le0$ and $g$ decreases — (I) is FALSE. On $\left(\dfrac\pi4,\dfrac\pi2\right)$, $\tan x>1$, so $g'\ge0$ and $g$ increases — (II) is FALSE.

Answer: neither statement is true; the truth is the exact opposite of both.

Check with a concrete $f$. Take $a=1$ and $f(x)=x^2$, which satisfies $f''=2>0$ and $f'(0)=0$. Then $g(x)=(\tan x-1)^4$, which takes the values $0.404,\ 0.042,\ 0.00062$ at $x=0.2,0.5,0.7$ (falling) and $0.0046,\ 6.11,\ 529.9$ at $x=0.9,1.2,1.4$ (rising). Exactly as proved.

Type 5 — Peeling an unknown $f$ off an inequality

This is the disguise JEE likes most. You are told only that some unnamed $f$ is strictly monotonic, and the entire problem is to convert an inequality in $f$ into an inequality in the arguments:

• $f$ strictly increasing: $f(u)<f(v)\iff u<v$.
• $f$ strictly decreasing: $f(u)<f(v)\iff u>v$ — the inequality flips.

Example 7 (JEE Main 2026, 08 Apr Shift 2). Let $\dfrac{x^2}{f(a^2+7a+3)}+\dfrac{y^2}{f(3a+15)}=1$ represent an ellipse with major axis along the $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb R$. If the set of all possible $a$ is $\mathbb R-[\alpha,\beta]$, find $\alpha^2+\beta^2$.

▸ SHOW SOLUTION

Step 1: Turn the geometry into an inequality. For $\dfrac{x^2}{A}+\dfrac{y^2}{B}=1$ with $A,B>0$, the major axis lies along $y$ exactly when $B>A$:

$f(3a+15)>f\left(a^2+7a+3\right).$

Both denominators are positive because $f$ is a positive function, so this really is an ellipse.

Step 2: Peel off $f$. $f$ is strictly decreasing, so comparing the inputs reverses the inequality:

$3a+15<a^2+7a+3.$

Step 3: Solve.

$0<a^2+4a-12=(a+6)(a-2)\ \Longrightarrow\ a>2\ \text{ or }\ a<-6.$

Step 4: Match the stated form. The set is $\mathbb R-[-6,2]$, so $\alpha=-6$, $\beta=2$.

$\alpha^2+\beta^2=36+4=\boxed{40}$

Check. At $a=3$: inputs $24$ and $33$, and $24<33$, so a decreasing $f$ gives $f(24)>f(33)$ — the $y^2$ denominator is indeed larger. At $a=0$, inside the excluded interval: inputs $15$ and $3$, so $f(15)<f(3)$ and the major axis would be along $x$. Correct.

Example 8 (JEE Main 2026, 05 Apr Shift 1). The system $x+2y+tz=0$, $6x+y+5tz=0$, $3x+t^2y+f(t)z=0$ has infinitely many solutions for every $t\in\mathbb R$, where $f$ is differentiable. What can be said about $f$?

▸ SHOW SOLUTION

Step 1: Determinant condition. A homogeneous system has non-trivial solutions exactly when

$\begin{vmatrix}1&2&t\\ 6&1&5t\\ 3&t^2&f(t)\end{vmatrix}=0.$

Step 2: Expand along the first row.

$1\big(f(t)-5t^3\big)-2\big(6f(t)-15t\big)+t\big(6t^2-3\big)=0.$

Step 3: Collect.

$-11f(t)+t^3+27t=0\ \Longrightarrow\ f(t)=\frac{t^3+27t}{11}.$

Step 4: Differentiate.

$f'(t)=\frac{3t^2+27}{11}\ \ge\ \frac{27}{11}>0\quad\text{for every real }t.$

Step 5: Conclude. $f$ is strictly increasing on $\mathbb R$ — in particular not constant, and with no critical point.

Type 6 — Proving an inequality, and counting roots

To prove $g(x)<h(x)$ on an interval, set $\varphi=h-g$, show $\varphi'>0$, and evaluate $\varphi$ at one convenient endpoint. Monotonicity then carries that single value across the whole interval.

Example 9. Prove that $\log_e(1+x)<x$ for every $x>0$.

▸ SHOW SOLUTION

Step 1: Difference function. $\varphi(x)=x-\log_e(1+x)$ on $[0,\infty)$.

Step 2: Differentiate.

$\varphi'(x)=1-\frac{1}{1+x}=\frac{x}{1+x}.$

Step 3: Sign. For $x>0$ both $x>0$ and $1+x>0$, so $\varphi'>0$; and $\varphi$ is continuous at $0$. So $\varphi$ is strictly increasing on $[0,\infty)$.

Step 4: Endpoint. $\varphi(0)=0-\log_e1=0$.

Step 5: Conclude. Strictly increasing from $0$ gives $\varphi(x)>0$ for $x>0$, i.e. $\log_e(1+x)<x$. $\blacksquare$

Check. At $x=1$: $\log_e2\approx0.693<1$. At $x=0.01$: $\log_e(1.01)\approx0.00995<0.01$.

Running the same recipe on $\log_e(1+x)-\dfrac{x}{1+x}$ gives the companion bound, and together they make the two-sided estimate worth memorising:

$\frac{x}{1+x}<\log_e(1+x)<x\qquad (x>0).$

For roots, monotonicity gives uniqueness and the intermediate value theorem gives existence: if $f$ is strictly monotonic on $I$, then $f(x)=k$ has at most one root there; add continuity and a sign change and it has exactly one.

Example 10. Show that $x^3+3x-7=0$ has exactly one real root, and locate it between consecutive integers.

▸ SHOW SOLUTION

Step 1: Monotonicity. $f'(x)=3x^2+3=3(x^2+1)\ge3>0$, so $f$ is strictly increasing on $\mathbb R$ — at most one root.

Step 2: Existence. $f(1)=1+3-7=-3<0$ and $f(2)=8+6-7=7>0$. The sign changes, so a root lies in $(1,2)$.

Step 3: Combine. Exactly one real root, and it lies in $(1,2)$.

Check. $f(1.4)=-0.056<0$ and $f(1.45)\approx0.399>0$, so the root sits near $1.41$.

Quick recall

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Work the linked chapters on doMath to see these patterns in past-paper form, and read the JEE Main 2026 shift-wise analyses to see how many marks Application of Derivatives actually carried this year.