Increasing and Decreasing Functions

What Does "Increasing" Mean?

Look at the graph of $f(x) = x^2$. To the right of the origin, as you move from left to right, the height of the graph continuously rises — the function is increasing for $x > 0$. To the left of the origin, moving left to right, the height continuously falls — the function is decreasing for $x < 0$.

Let's make this precise. Let $I$ be an interval contained in the domain of a real valued function $f$. Then $f$ is said to be:

1. Increasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) < f(x_2)$ for all $x_1, x_2 \in I$
2. Decreasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) > f(x_2)$ for all $x_1, x_2 \in I$
3. Constant on $I$ if $f(x) = c$ for all $x \in I$, where $c$ is a constant
4. Strictly increasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) < f(x_2)$
5. Strictly decreasing on $I$ if $x_1 < x_2$ in $I \Rightarrow f(x_1) > f(x_2)$

A function can also be increasing or decreasing at a point $x_0$: this means there is some open interval $I$ containing $x_0$ on which the function is increasing (or decreasing).

Key Point: Bigger input $\Rightarrow$ bigger output means increasing. Bigger input $\Rightarrow$ smaller output means decreasing. And a function like $x^2$ can be decreasing on one interval and increasing on another — always name the interval.

[JEE Tip] Some books (and older NCERT editions) define "increasing" with $f(x_1) \leq f(x_2)$ (non-decreasing) and reserve $<$ for "strictly increasing". In JEE, "monotonically increasing" usually allows flat stretches. Read the question's convention carefully — but for Board answers, follow the current NCERT definitions above.

The First Derivative Test for Monotonicity

Checking $f(x_1) < f(x_2)$ for every pair of points is impractical. The derivative gives a far better tool.

Theorem 1. Let $f$ be continuous on $[a, b]$ and differentiable on the open interval $(a, b)$. Then:

• (a) $f$ is increasing in $[a, b]$ if $f'(x) > 0$ for each $x \in (a, b)$
• (b) $f$ is decreasing in $[a, b]$ if $f'(x) < 0$ for each $x \in (a, b)$
• (c) $f$ is a constant function in $[a, b]$ if $f'(x) = 0$ for each $x \in (a, b)$

Why it works (proof sketch): Take any $x_1 < x_2$ in $[a, b]$. By the Mean Value Theorem there is a point $c$ between them with

$$f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$$

If $f'(c) > 0$, the right side is positive, so $f(x_2) > f(x_1)$ — that is exactly "increasing". The tangent-slope sign controls the function's direction.

Key Point: Positive slope everywhere $\Rightarrow$ graph climbs. Negative slope everywhere $\Rightarrow$ graph falls. The test needs the sign of $f'$ on the open interval; continuity extends the conclusion to the closed interval's endpoints.

[JEE Important] $f'$ vanishing at isolated points does no harm: $f(x) = x^3$ has $f'(0) = 0$, yet $x^3$ is strictly increasing on all of $\mathbb{R}$ because $f'(x) = 3x^2 > 0$ everywhere else. "$f'(x) \geq 0$ with equality only at isolated points" still gives strict increase — a favourite JEE conceptual trap.

The Working Method: Sign Analysis on the Number Line

Most exam questions say: "Find the intervals in which $f$ is increasing or decreasing." Here is the standard recipe:

1. Differentiate: compute $f'(x)$.
2. Solve $f'(x) = 0$: these critical values split the real line (or the given domain) into disjoint intervals.
3. Factorise $f'$ and test the sign of each factor in each interval (a quick sign table).
4. Conclude: $f' > 0 \Rightarrow$ increasing on that interval; $f' < 0 \Rightarrow$ decreasing.

Take $f(x) = 4x^3 - 6x^2 - 72x + 30$:

$$f'(x) = 12x^2 - 12x - 72 = 12(x - 3)(x + 2)$$

So $f'(x) = 0$ at $x = -2$ and $x = 3$, giving three intervals:

Note that $f$ is increasing on $(-\infty, -2)$ and on $(3, \infty)$, decreasing on $(-2, 3)$ — but it is neither increasing nor decreasing on $\mathbb{R}$ as a whole.

Key Point: Never write "increasing on $(-\infty, -2) \cup (3, \infty)$" as one monotonic claim. Monotonicity is interval-by-interval; the union statement is false (compare $f(-2)$ with $f(3)$ here). State the intervals separately.

[Board Important] Present the sign table in your Board answer — it earns method marks even if an interval slips.

Trigonometric Functions Need Extra Care

For trig functions the critical values come from solving equations like $\cos 3x = 0$ within the given domain.

Example pattern 1: $f(x) = \cos x$ has $f'(x) = -\sin x$. On $(0, \pi)$, $\sin x > 0$, so $f'(x) < 0$: cosine is decreasing on $(0, \pi)$. On $(\pi, 2\pi)$, $\sin x < 0$, so $f'(x) > 0$: cosine is increasing there. Over the whole of $(0, 2\pi)$ it is neither.

Example pattern 2: $f(x) = \sin x + \cos x$ on $[0, 2\pi]$. Then $f'(x) = \cos x - \sin x$, which vanishes where $\tan x = 1$, i.e. at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. Sign analysis gives: increasing on $\left[0, \frac{\pi}{4}\right)$ and $\left(\frac{5\pi}{4}, 2\pi\right]$, decreasing on $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.

Key Point: Always solve $f'(x) = 0$ within the stated domain — $\cos 3x = 0$ on $\left(0, \frac{\pi}{2}\right)$ gives $3x = \frac{\pi}{2}, \frac{3\pi}{2}$, i.e. $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$, because $3x$ ranges over $\left(0, \frac{3\pi}{2}\right)$.

[JEE Tip] A quick alternative for $\sin x + \cos x$: write it as $\sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$. It increases exactly where the shifted sine increases — the same intervals appear with almost no computation.

Parameters and Clever Algebraic Forms

Two patterns that Boards and JEE both love:

1. Finding parameter values

For what values of $a$ is $f(x) = x^2 + ax + 1$ increasing on $[1, 2]$?

Here $f'(x) = 2x + a$. We need $f'(x) > 0$ for all $x \in (1, 2)$. Since $2x + a$ is smallest as $x \to 1^+$, the condition is $2(1) + a \geq 0$, i.e. $a \geq -2$. (At $a = -2$, $f'(x) = 2x - 2 > 0$ for every $x$ in the open interval $(1,2)$, so it qualifies.)

2. Showing a function is increasing via a perfect square

Show $y = \log(1+x) - \frac{2x}{2+x}$ is increasing for $x > -1$.

$$\frac{dy}{dx} = \frac{1}{1+x} - \frac{4}{(2+x)^2} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$$

For $x > -1$ the denominator is positive and $x^2 \geq 0$, so $\frac{dy}{dx} \geq 0$ (zero only at the isolated point $x = 0$) — increasing throughout the domain.

3. The $x + \frac{1}{x}$ family

For $f(x) = x + \frac{1}{x}$, $f'(x) = 1 - \frac{1}{x^2} = \frac{(x-1)(x+1)}{x^2}$. So $f' > 0$ exactly when $|x| > 1$: the function is increasing on any interval disjoint from $[-1, 1]$, and decreasing on $(-1, 0)$ and $(0, 1)$.

[JEE Important] When $f'$ can be written as $\frac{(\text{perfect square})}{(\text{positive})}$ or factored into simple linear factors, monotonicity questions become sign-reading exercises. Practise forcing $f'$ into such forms — it is the single most useful algebraic skill for this topic.

Solved Examples

Example 1: A linear function, straight from the definition

Show that the function given by $f(x) = 7x - 3$ is increasing on $\mathbb{R}$.

Solution:

1. Take any two reals with $x_1 < x_2$.
2. Multiply by 7 (positive, so inequality keeps direction): $7x_1 < 7x_2$.
3. Subtract 3: $7x_1 - 3 < 7x_2 - 3$, i.e. $f(x_1) < f(x_2)$.

Final Answer: By the definition, $f$ is (strictly) increasing on $\mathbb{R}$.

Takeaway: For simple functions the definition itself is the fastest route — no derivative needed.

Example 2: Exponential growth is always increasing

Show that the function given by $f(x) = e^{2x}$ is increasing on $\mathbb{R}$.

Solution:

1. Differentiate: $f'(x) = 2e^{2x}$.
2. Sign: $e^{2x} > 0$ for every real $x$, so $f'(x) = 2e^{2x} > 0$ on all of $\mathbb{R}$.
3. Apply Theorem 1: $f' > 0$ everywhere $\Rightarrow$ $f$ is increasing on $\mathbb{R}$.

Final Answer: $e^{2x}$ is increasing on $\mathbb{R}$.

Takeaway: Exponentials with positive coefficients in the exponent are always increasing — their derivative is a positive multiple of themselves.

Example 3: When the derivative hides a perfect square

Show that $f(x) = x^3 - 3x^2 + 4x$, $x \in \mathbb{R}$, is increasing on $\mathbb{R}$.

Solution:

1. Differentiate: $f'(x) = 3x^2 - 6x + 4$.
2. Complete the square: $f'(x) = 3(x^2 - 2x + 1) + 1 = 3(x-1)^2 + 1$.
3. Sign: $3(x-1)^2 \geq 0$, so $f'(x) \geq 1 > 0$ for every $x$.

Final Answer: $f'(x) > 0$ in every interval of $\mathbb{R}$, so $f$ is increasing on $\mathbb{R}$.

Takeaway: When $f'$ is a quadratic with negative discriminant (here $36 - 48 < 0$), it never changes sign — completing the square shows this instantly.

Example 4: Cosine on different intervals

Prove that $f(x) = \cos x$ is (a) decreasing in $(0, \pi)$, (b) increasing in $(\pi, 2\pi)$, (c) neither increasing nor decreasing in $(0, 2\pi)$.

Solution:

1. Differentiate: $f'(x) = -\sin x$.
2. (a) For $x \in (0, \pi)$: $\sin x > 0 \Rightarrow f'(x) < 0 \Rightarrow f$ is decreasing in $(0, \pi)$.
3. (b) For $x \in (\pi, 2\pi)$: $\sin x < 0 \Rightarrow f'(x) > 0 \Rightarrow f$ is increasing in $(\pi, 2\pi)$.
4. (c) Since $f$ decreases on part of $(0, 2\pi)$ and increases on another part, it is neither increasing nor decreasing on $(0, 2\pi)$.

Final Answer: As proved above.

Takeaway: A function's behaviour is interval-specific — one interval where it rises and one where it falls means no overall monotonicity.

Example 5: A quadratic — one critical point

Find the intervals in which $f(x) = x^2 - 4x + 6$ is (a) increasing (b) decreasing.

Solution:

1. Differentiate: $f'(x) = 2x - 4$.
2. Critical value: $f'(x) = 0 \Rightarrow x = 2$, splitting $\mathbb{R}$ into $(-\infty, 2)$ and $(2, \infty)$.
3. Signs: On $(-\infty, 2)$: $f'(x) < 0$ (decreasing). On $(2, \infty)$: $f'(x) > 0$ (increasing).

Final Answer: $f$ is decreasing on $(-\infty, 2)$ and increasing on $(2, \infty)$.

Takeaway: A parabola opening upwards always decreases to the left of its vertex and increases to the right.

Example 6: A cubic — two critical points

Find the intervals in which $f(x) = 4x^3 - 6x^2 - 72x + 30$ is (a) increasing (b) decreasing.

Solution:

1. Differentiate and factorise: $f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x - 3)(x + 2)$.
2. Critical values: $x = -2$ and $x = 3$, giving intervals $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$.
3. Sign table: On $(-\infty, -2)$: $(-)(-) > 0$. On $(-2, 3)$: $(-)(+) < 0$. On $(3, \infty)$: $(+)(+) > 0$.

Final Answer: $f$ is increasing on $(-\infty, -2)$ and $(3, \infty)$; decreasing on $(-2, 3)$.

Takeaway: Factorise $f'$ completely, then read signs interval by interval. This is the model answer format for a 3-mark Board question.

Example 7: A trig function with a compressed argument

Find the intervals in which $f(x) = \sin 3x$, $x \in \left[0, \frac{\pi}{2}\right]$, is (a) increasing (b) decreasing.

Solution:

1. Differentiate: $f'(x) = 3\cos 3x$.
2. Solve $f'(x) = 0$ in the domain: As $x$ runs over $\left[0, \frac{\pi}{2}\right]$, $3x$ runs over $\left[0, \frac{3\pi}{2}\right]$. So $\cos 3x = 0$ gives $3x = \frac{\pi}{2}, \frac{3\pi}{2}$, i.e. $x = \frac{\pi}{6}, \frac{\pi}{2}$.
3. Signs: For $x \in \left(0, \frac{\pi}{6}\right)$: $3x \in \left(0, \frac{\pi}{2}\right)$, so $\cos 3x > 0$, $f' > 0$. For $x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right)$: $3x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, so $\cos 3x < 0$, $f' < 0$.
4. Include endpoints by continuity.

Final Answer: Increasing on $\left[0, \frac{\pi}{6}\right]$, decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

Takeaway: Track the range of the inner argument ($3x$ here), not just $x$ — that's where students most often lose a critical point.

Example 8: $\sin x + \cos x$ on a full period

Find the intervals in which $f(x) = \sin x + \cos x$, $0 \leq x \leq 2\pi$, is increasing or decreasing.

Solution:

1. Differentiate: $f'(x) = \cos x - \sin x$.
2. Solve $f'(x) = 0$: $\sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4}$ in $[0, 2\pi]$.
3. Sign analysis over the three intervals: $f' > 0$ on $\left[0, \frac{\pi}{4}\right)$ (e.g. at $x=0$, $f' = 1 > 0$); $f' < 0$ on $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$ (e.g. at $x = \pi$, $f' = -1 < 0$); $f' > 0$ on $\left(\frac{5\pi}{4}, 2\pi\right]$ (e.g. at $x = \frac{3\pi}{2}$, $f' = 1 > 0$).

Final Answer: Increasing on $\left[0, \frac{\pi}{4}\right]$ and $\left[\frac{5\pi}{4}, 2\pi\right]$; decreasing on $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.

Takeaway: Test one convenient point inside each interval to fix the sign of $f'$ — quicker and safer than reasoning abstractly.

Example 9: Standard Board cubic

Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is (a) increasing (b) decreasing.

Solution:

1. Differentiate and factorise: $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2)$.
2. Critical values: $x = -2, 3$.
3. Signs: Positive on $(-\infty, -2)$, negative on $(-2, 3)$, positive on $(3, \infty)$.

Final Answer: Increasing on $(-\infty, -2)$ and $(3, \infty)$; decreasing on $(-2, 3)$.

Takeaway: This exact structure — cubic with two critical points — is the most repeated monotonicity question in Board papers.

Example 10: A product needing careful factoring

Find the values of $x$ for which $y = [x(x-2)]^2$ is an increasing function.

Solution:

1. Rewrite: $y = (x^2 - 2x)^2$.
2. Differentiate (Chain Rule): $\frac{dy}{dx} = 2(x^2 - 2x)(2x - 2) = 4x(x - 2)(x - 1)$.
3. Critical values: $x = 0, 1, 2$ — four intervals.
4. Sign table: On $(-\infty, 0)$: $(-)(-)(-) < 0$. On $(0, 1)$: $(+)(-)(-) > 0$. On $(1, 2)$: $(+)(-)(+) < 0$. On $(2, \infty)$: $(+)(+)(+) > 0$.

Final Answer: $y$ is increasing for $0 < x < 1$ and $x > 2$ (i.e. on $(0,1)$ and $(2, \infty)$).

Takeaway: With three critical points you get four intervals — a sign table keeps the bookkeeping error-free.

Example 11: Increasing via a perfect square (a classic proof)

Show that $y = \log(1+x) - \frac{2x}{2+x}$, $x > -1$, is an increasing function of $x$ throughout its domain.

Solution:

1. Differentiate: $\frac{dy}{dx} = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$.
2. Combine over a common denominator: $\frac{dy}{dx} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$.
3. Sign: For $x > -1$: numerator $x^2 \geq 0$ and denominator $> 0$, so $\frac{dy}{dx} \geq 0$, vanishing only at the single point $x = 0$.

Final Answer: $y$ is increasing throughout $(-1, \infty)$.

Takeaway: The whole battle is algebraic simplification of $y'$. When the numerator collapses to a perfect square, the sign question answers itself.

Example 12: Finding parameter values

For what values of $a$ is $f(x) = x^2 + ax + 1$ increasing on $[1, 2]$?

Solution:

1. Differentiate: $f'(x) = 2x + a$.
2. Condition: We need $f'(x) > 0$ for all $x \in (1, 2)$.
3. Worst case: $2x + a$ is smallest near $x = 1$, so we need $2 + a \geq 0$, i.e. $a \geq -2$. (Check $a = -2$: $f'(x) = 2x - 2 > 0$ for every $x$ in the open interval $(1,2)$ — it works.)

Final Answer: $a \geq -2$.

Takeaway: For parameter problems, force the inequality at the worst point of the interval. Check boundary parameter values against the open interval before accepting or rejecting them.

Example 13: Increasing away from $[-1, 1]$ — a JEE favourite

Let $I$ be any interval disjoint from $[-1, 1]$. Prove that $f(x) = x + \frac{1}{x}$ is increasing on $I$.

Solution:

1. Differentiate: $f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} = \frac{(x-1)(x+1)}{x^2}$.
2. Sign on $I$: If $I$ is disjoint from $[-1, 1]$, then every $x \in I$ has $|x| > 1$, so $x^2 - 1 > 0$; also $x^2 > 0$.
3. Conclude: $f'(x) > 0$ for all $x \in I$, so $f$ is increasing on $I$.

Final Answer: Proved — $f$ is increasing on every interval disjoint from $[-1, 1]$.

Takeaway: $x + \frac{1}{x}$ decreases on $(-1, 0)$ and $(0, 1)$ and increases outside $[-1,1]$. Remember its shape — it appears constantly in JEE maxima-minima and inequality problems.