Chapter 6 · Application of Derivatives

6.2 Increasing and Decreasing Functions

Where is a graph going uphill and where downhill? Read the sign of $f'(x)$: positive means rising, negative means falling. A sign chart of $f'$ splits the real line into intervals of increase and decrease.

A cardiologist looks at a patient's blood-pressure trace and asks a simple question about each stretch: is it rising or falling? A shopkeeper asks the same question about weekly sales, a hydrologist about a river's level during the monsoon. For a quantity given by a formula, plotting thousands of points is slow and can still mislead you — a graph that looks flat may be quietly rising. Calculus answers the question exactly. In Section 6.1 we read $f'(x)$ as a rate of change; a positive rate means the quantity is going up, a negative rate means it is going down.

This section makes that precise. We first define what it means for a function to be increasing or decreasing on an interval (and the stronger idea of strictly increasing or decreasing), then see why the sign of $f'$ decides the matter (the derivative test), and finally learn the routine that Board papers ask for again and again: factorise $f'(x)$, make a sign chart, and read off the intervals.

Increasing and Decreasing Functions

Think of walking along a graph from left to right. On some stretches you climb, on others you descend, and on flat stretches you neither climb nor descend. We describe this by comparing the outputs of any two inputs.

📘 Definition 6.3 — Increasing and decreasing on an interval
Let $I$ be an interval contained in the domain of a real function $f$. Then $f$ is
  • increasing on $I$ if $x_1\lt x_2$ in $I$ $\Rightarrow f(x_1)\le f(x_2)$;
  • strictly increasing on $I$ if $x_1\lt x_2$ in $I$ $\Rightarrow f(x_1)\lt f(x_2)$;
  • decreasing on $I$ if $x_1\lt x_2$ in $I$ $\Rightarrow f(x_1)\ge f(x_2)$;
  • strictly decreasing on $I$ if $x_1\lt x_2$ in $I$ $\Rightarrow f(x_1)\gt f(x_2)$.
A function that is increasing on $I$ or decreasing on $I$ is called monotonic on $I$.

"Increasing" allows flat stretches ($f(x_1)=f(x_2)$ is permitted); "strictly increasing" does not. So a constant function is both increasing and decreasing on every interval, but neither strictly. Every strictly increasing function is increasing, not conversely.

📘 Definition 6.4 — Increasing or decreasing at a point
Let $x_0$ be a point in the domain of $f$. Then $f$ is strictly increasing at $x_0$ if there is an open interval $I$ containing $x_0$ on which $f$ is strictly increasing, and strictly decreasing at $x_0$ if there is such an interval on which $f$ is strictly decreasing.
✏️ Example 1 — Using only the definition
Without derivatives, show that (a) $f(x)=3x+5$ is strictly increasing on $\mathbb R$; (b) $g(x)=x^2$ is strictly decreasing on $(-\infty,0]$ and strictly increasing on $[0,\infty)$, but neither increasing nor decreasing on $\mathbb R$; (c) $h(x)=-2x+1$ is strictly decreasing on $\mathbb R$.
(a) If $x_1\lt x_2$, then $3x_1\lt3x_2$ (multiplying by a positive number keeps the inequality), so $3x_1+5\lt3x_2+5$, i.e. $f(x_1)\lt f(x_2)$.
(b) On $[0,\infty)$: if $0\le x_1\lt x_2$ then $x_2^2-x_1^2=(x_2-x_1)(x_2+x_1)\gt0$, because both factors are positive. So $g(x_1)\lt g(x_2)$. On $(-\infty,0]$: if $x_1\lt x_2\le0$ then $x_2-x_1\gt0$ but $x_2+x_1\lt0$, so $x_2^2-x_1^2\lt0$: strictly decreasing. On all of $\mathbb R$: $-1\lt0$ with $g(-1)=1\gt g(0)=0$ rules out "increasing", and $0\lt1$ with $g(0)\lt g(1)$ rules out "decreasing".
(c) If $x_1\lt x_2$ then $-2x_1\gt-2x_2$ (multiplying by a negative number reverses the inequality), so $h(x_1)\gt h(x_2)$.
Lesson In (b) one counterexample pair is enough to show a function is not monotonic on an interval; proving that it is monotonic needs an argument for every pair. The derivative test below does this work for us.

The Derivative Test for Increasing and Decreasing Functions

Suppose $f'(x)\gt0$ at every point of an interval. Then at every point the graph is climbing: over any tiny step to the right, the change in $f$ is about $f'(x)$ times the step, which is positive. Adding up these tiny climbs from $x_1$ to $x_2$ gives a total climb $f(x_2)-f(x_1)\gt0$. This is the idea behind the most-used theorem of the chapter.

📐 Theorem 6.2 — Derivative test for monotonicity
Let $f$ be continuous on $[a,b]$ and differentiable on the open interval $(a,b)$.
  1. If $f'(x)\gt0$ for every $x\in(a,b)$, then $f$ is strictly increasing on $[a,b]$.
  2. If $f'(x)\lt0$ for every $x\in(a,b)$, then $f$ is strictly decreasing on $[a,b]$.
  3. If $f'(x)=0$ for every $x\in(a,b)$, then $f$ is constant on $[a,b]$.
The same conclusions hold on any interval (open, half-open or unbounded, such as $\mathbb R$) if the hypotheses hold at every interior point and $f$ is continuous on the interval.

Why it is true. The climbing argument above is the heart of the matter: a graph whose slope is positive at every point cannot come back down. A complete proof needs an additional theorem of calculus that is not part of the Board syllabus, so we accept Theorem 6.2 on the strength of this argument — just as you accepted the limit $\lim\limits_{x\to0}\tfrac{\sin x}{x}=1$ from a picture in Class 11.

Two refinements are used constantly:

Some quick applications (throughout this chapter $\log x$ means the natural logarithm $\log_e x$):

Finding the Intervals of Increase and Decrease

Most functions rise on some intervals and fall on others. The places where they can switch are the points where $f'(x)=0$ (or where $f'$ fails to exist). Between two consecutive such points, a continuous $f'$ cannot change sign — to go from positive to negative it would have to pass through $0$ — so one test value in each piece tells you the sign on the whole piece.

🧭 Problem-solving strategy — the sign chart of $f'$
  1. Find the domain of $f$ and compute $f'(x)$. Factorise it as far as you can.
  2. Solve $f'(x)=0$ and note any points where $f'$ does not exist. Mark these points on a number line; they split the domain into intervals.
  3. Find the sign of $f'$ on each interval — from the signs of the factors, or by one test value.
  4. Read off: $f'\gt0$ ⇒ strictly increasing; $f'\lt0$ ⇒ strictly decreasing. Include an end point in an interval when $f$ is continuous there (Theorem 6.2 allows closed intervals).
  5. Check with a value or two of $f$ itself.
✏️ Example 2 — A cubic
Find the intervals on which $f(x)=2x^3-9x^2+12x+1$ is strictly increasing and strictly decreasing.
Derivative $f'(x)=6x^2-18x+12$ $=6(x^2-3x+2)$ $=6(x-1)(x-2)$.
Split $f'(x)=0$ at $x=1$ and $x=2$, giving the intervals $(-\infty,1)$, $(1,2)$, $(2,\infty)$.
Signs
interval$x-1$$x-2$$f'(x)$$f$
$(-\infty,1)$$-$$-$$+$strictly increasing
$(1,2)$$+$$-$$-$strictly decreasing
$(2,\infty)$$+$$+$$+$strictly increasing
Answer $f$ is strictly increasing on $(-\infty,1]$ and on $[2,\infty)$, and strictly decreasing on $[1,2]$.
Check $f(0)=1$, $f(1)=6$, $f(2)=5$, $f(3)=10$: up, down, up ✓. Note that $f$ is not increasing on $(-\infty,1]\cup[2,\infty)$ as a whole: $f(1)=6\gt f(2)=5$. Always list the intervals separately.
Interactive 1 — The Increasing / Decreasing Map

Choose a function. The top panel is its graph, painted green where $f'(x)\gt0$ and red where $f'(x)\lt0$. The strip underneath is the sign chart of $f'$: the zeros of $f'$ (purple) cut the line into intervals, each labelled $+$ or $-$. Press Play to send the point along the curve, or drag it with the slider; its short dark arrow shows the slope $f'(x)$.

$x$—
$f(x)$—
$f'(x)$—
here $f$ is—
✏️ Example 3 — A quartic with three turning points
Find the intervals of increase and decrease of $f(x)=x^4-2x^2$.
Derivative $f'(x)=4x^3-4x=4x(x-1)(x+1)$, which is $0$ at $x=-1,0,1$.
Signs Test values: $f'(-2)=-24\lt0$; $f'\left(-\tfrac12\right)=\tfrac32\gt0$; $f'\left(\tfrac12\right)=-\tfrac32\lt0$; $f'(2)=24\gt0$.
Answer Strictly decreasing on $(-\infty,-1]$ and $[0,1]$; strictly increasing on $[-1,0]$ and $[1,\infty)$.
Check $f(-2)=8$, $f(-1)=-1$, $f(0)=0$, $f(1)=-1$, $f(2)=8$: down, up, down, up ✓ — a "W" shape.
✏️ Example 4 — A trigonometric function on a bounded interval
Find the intervals in $[0,2\pi]$ on which $f(x)=\sin x+\cos x$ is strictly increasing or strictly decreasing.
Derivative $f'(x)=\cos x-\sin x$. It is $0$ when $\tan x=1$, i.e. at $x=\tfrac\pi4$ and $x=\tfrac{5\pi}4$ in $[0,2\pi]$.
Signs $f'(0)=1\gt0$; $f'\left(\tfrac\pi2\right)=-1\lt0$; $f'\left(\tfrac{3\pi}2\right)=0-(-1)=1\gt0$.
Answer Strictly increasing on $\left[0,\tfrac\pi4\right]$ and $\left[\tfrac{5\pi}4,2\pi\right]$; strictly decreasing on $\left[\tfrac\pi4,\tfrac{5\pi}4\right]$.
Check $\sin x+\cos x=\sqrt2\sin\left(x+\tfrac\pi4\right)$, which climbs to $\sqrt2$ at $x=\tfrac\pi4$ and falls to $-\sqrt2$ at $x=\tfrac{5\pi}4$ ✓.
✏️ Example 5 — Choosing a constant
Find all real $k$ for which $f(x)=x^3+kx^2+3x+5$ is strictly increasing on $\mathbb R$.
Condition $f'(x)=3x^2+2kx+3$. We need $f'(x)\ge0$ for all $x$, with equality only at isolated points. A quadratic with positive leading coefficient is never negative exactly when its discriminant is $\le0$.
Discriminant $(2k)^2-4(3)(3)=4k^2-36\le0$ $\iff k^2\le9$ $\iff-3\le k\le3$.
End cases At $k=\pm3$, $f'(x)=3x^2\pm6x+3=3(x\pm1)^2$, which is $0$ at a single point only, so $f$ is still strictly increasing. For $|k|\gt3$, $f'$ is negative between its two roots and $f$ decreases there.
Answer $k\in[-3,3]$. Explore it in Interactive 2.
Interactive 2 — When Is a Cubic Increasing Everywhere?

Here $f(x)=x^3+kx^2+3x+5$ (left) and its derivative $f'(x)=3x^2+2kx+3$ (right). Move $k$, or press Play to sweep it from $-5$ to $5$. Whenever the parabola $y=f'(x)$ dips below the axis, the cubic has a falling stretch (painted red). The parabola just touches the axis at $k=\pm3$, and stays on or above it exactly for $-3\le k\le3$.

$k$—
discriminant $4k^2-36$—
least value of $f'$—
strictly increasing on ℝ?—
✏️ Example 6 — Proving an inequality by monotonicity
Prove that $\sin x\lt x$ for every $x\gt0$.
Idea Turn the inequality into "a function is positive": let $g(x)=x-\sin x$. We must show $g(x)\gt0$ for $x\gt0$.
Monotonic $g'(x)=1-\cos x\ge0$, and $g'(x)=0$ only where $\cos x=1$, i.e. at the isolated points $x=0,2\pi,4\pi,\dots$ In any bounded interval there are only finitely many of them, so $g$ is strictly increasing on $[0,\infty)$.
Conclude For $x\gt0$: $g(x)\gt g(0)=0$, i.e. $x-\sin x\gt0$, so $\sin x\lt x$. $\square$
Check $x=1$: $\sin1\approx0.8415\lt1$ ✓; $x=0.01$: $\sin0.01\approx0.0099998\lt0.01$ ✓.
⚠️ Common mistakes when studying alone

1. Joining intervals with ∪. In Example 2, "$f$ is increasing on $(-\infty,1]\cup[2,\infty)$" is wrong as a single statement, because $f(1)\gt f(2)$. Say "on $(-\infty,1]$ and on $[2,\infty)$".

2. Forgetting the domain. For $\log x$ or $\dfrac1x$, the sign chart lives only on the domain. $\dfrac1x$ is strictly decreasing on $(-\infty,0)$ and on $(0,\infty)$, but not on $\mathbb R-\{0\}$: $-1\lt1$ yet $\tfrac1{-1}\lt\tfrac11$.

3. Thinking $f'(c)=0$ forces a change. $x^3$ and $x^4/4-x^3$ (at $x=0$) have $f'=0$ without $f'$ changing sign. Always test both sides.

4. Sign errors in factors. Write the factors in the form $(x-a)$ and use a table; with a negative leading coefficient (e.g. $f'=-3(x-1)(x+2)$), every sign flips.

5. Mixing up "increasing" and "strictly increasing". Board questions usually say "strictly"; for a function like $x^3$, the answer "strictly increasing on $\mathbb R$" needs the isolated-zeros remark after Theorem 6.2.

Exercises

I
Show that each function has the stated property.
  • 1 $f(x)=7x-3$ is strictly increasing on $\mathbb R$.
  • 2 $f(x)=e^{3x}$ is strictly increasing on $\mathbb R$.
  • 3 $f(x)=\cos x$ is strictly decreasing on $(0,\pi)$ and strictly increasing on $(\pi,2\pi)$.
  • 4 $f(x)=x^2-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$.
  • 5 $f(x)=\log(2x+1)$ is strictly increasing on $\left(-\tfrac12,\infty\right)$.
  • 6 $f(x)=x^3-3x^2+3x-100$ is strictly increasing on $\mathbb R$.
  • 7 $f(x)=\tan^{-1}x$ is strictly increasing on $\mathbb R$.
  • 8 $f(x)=\dfrac1x$ is strictly decreasing on $(0,\infty)$ and on $(-\infty,0)$, but not on its whole domain.
1 $f'(x)=7\gt0$ for all $x$.
2 $f'(x)=3e^{3x}\gt0$ for all $x$.
3 $f'(x)=-\sin x$, which is $\lt0$ on $(0,\pi)$ (where $\sin x\gt0$) and $\gt0$ on $(\pi,2\pi)$ (where $\sin x\lt0$).
4 $f'(x)=2x-1$ is negative on $\left(-1,\tfrac12\right)$ and positive on $\left(\tfrac12,1\right)$, so $f$ decreases and then increases. Concretely $f(0)=1\gt f\left(\tfrac12\right)=\tfrac34\lt f(1)=1$.
5 Domain $2x+1\gt0$; $f'(x)=\dfrac{2}{2x+1}\gt0$ there.
6 $f'(x)=3x^2-6x+3=3(x-1)^2\ge0$, zero only at $x=1$: strictly increasing on $\mathbb R$ (isolated zero).
7 $f'(x)=\dfrac1{1+x^2}\gt0$ for all $x$ (Chapter 5).
8 $f'(x)=-\dfrac1{x^2}\lt0$ on each of $(-\infty,0)$ and $(0,\infty)$. But $-1\lt1$ and $f(-1)=-1\lt f(1)=1$, so $f$ is not decreasing on $\mathbb R-\{0\}$ (Theorem 6.2 needs an interval).
II
Find the intervals on which $f$ is strictly increasing and strictly decreasing.
  • 1 $f(x)=x^2-4x+5$
  • 2 $f(x)=10-6x-2x^2$
  • 3 $f(x)=2x^3-3x^2-36x+7$
  • 4 $f(x)=x^3-12x$
  • 5 $f(x)=\tfrac14x^4-x^3$
  • 6 $f(x)=xe^{-x}$
  • 7 $f(x)=\dfrac{x}{x^2+1}$
  • 8 $f(x)=x^2e^x$
  • 9 $f(x)=\sin x-\cos x$, $0\le x\le2\pi$
  • 10 $f(x)=x-2\log x$, $x\gt0$
1 $f'=2x-4$: decreasing on $(-\infty,2]$, increasing on $[2,\infty)$.
2 $f'=-6-4x$, positive for $x\lt-\tfrac32$: increasing on $\left(-\infty,-\tfrac32\right]$, decreasing on $\left[-\tfrac32,\infty\right)$.
3 $f'=6x^2-6x-36=6(x-3)(x+2)$: increasing on $(-\infty,-2]$ and $[3,\infty)$, decreasing on $[-2,3]$.
4 $f'=3x^2-12=3(x-2)(x+2)$: increasing on $(-\infty,-2]$ and $[2,\infty)$, decreasing on $[-2,2]$.
5 $f'=x^3-3x^2=x^2(x-3)$. The factor $x^2\ge0$ does not change sign at $0$, so $f'\lt0$ for $x\lt3$ ($x\ne0$) and $f'\gt0$ for $x\gt3$: decreasing on $(-\infty,3]$, increasing on $[3,\infty)$.
6 $f'=e^{-x}-xe^{-x}=(1-x)e^{-x}$, and $e^{-x}\gt0$: increasing on $(-\infty,1]$, decreasing on $[1,\infty)$.
7 $f'=\dfrac{(x^2+1)-2x^2}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}$: increasing on $[-1,1]$, decreasing on $(-\infty,-1]$ and $[1,\infty)$.
8 $f'=2xe^x+x^2e^x=x(x+2)e^x$: increasing on $(-\infty,-2]$ and $[0,\infty)$, decreasing on $[-2,0]$.
9 $f'=\cos x+\sin x=0\iff\tan x=-1$: $x=\tfrac{3\pi}4,\tfrac{7\pi}4$. $f'(0)=1\gt0$, $f'(\pi)=-1\lt0$, $f'\left(\tfrac{15\pi}8\right)\gt0$. Increasing on $\left[0,\tfrac{3\pi}4\right]$ and $\left[\tfrac{7\pi}4,2\pi\right]$; decreasing on $\left[\tfrac{3\pi}4,\tfrac{7\pi}4\right]$.
10 $f'=1-\dfrac2x=\dfrac{x-2}x$ with $x\gt0$: decreasing on $(0,2]$, increasing on $[2,\infty)$.
III
Parameters and inequalities.
  • 1 Find all values of $a$ for which $f(x)=ax+\cos x$ is strictly increasing on $\mathbb R$.
  • 2 Find all values of $k$ for which $f(x)=x^3-3kx^2+12x$ is strictly increasing on $\mathbb R$.
  • 3 Show that $f(x)=x^3-6x^2+15x-4$ is strictly increasing on $\mathbb R$.
  • 4 Prove that $e^x\gt1+x$ for every $x\gt0$.
  • 5 Prove that $\tan x\gt x$ for $0\lt x\lt\tfrac\pi2$.
  • 6 Show that $f(x)=\dfrac{\log x}{x}$ is strictly decreasing on $[e,\infty)$, and deduce that $3^{1/3}\gt4^{1/4}$.
  • 7 Find the intervals in $\left[0,\tfrac\pi2\right]$ on which $f(x)=\sin^4x+\cos^4x$ is strictly increasing and strictly decreasing.
1 $f'(x)=a-\sin x$. Need $a\ge\sin x$ for all $x$, i.e. $a\ge1$. At $a=1$, $f'=1-\sin x$ vanishes only at the isolated points $x=\tfrac\pi2+2n\pi$, so $f$ is still strictly increasing. Answer: $a\ge1$.
2 $f'=3x^2-6kx+12$; discriminant $36k^2-144\le0\iff|k|\le2$ (at $k=\pm2$, $f'=3(x\mp2)^2$ has one zero). Answer: $-2\le k\le2$.
3 $f'=3x^2-12x+15$ $=3\left[(x-2)^2+1\right]\ge3\gt0$.
4 $g(x)=e^x-1-x$ has $g'(x)=e^x-1\gt0$ for $x\gt0$, so $g$ is strictly increasing on $[0,\infty)$ and $g(x)\gt g(0)=0$ for $x\gt0$.
5 $g(x)=\tan x-x$ has $g'(x)=\sec^2x-1=\tan^2x\gt0$ on $\left(0,\tfrac\pi2\right)$; $g$ is continuous on $\left[0,\tfrac\pi2\right)$ with $g(0)=0$, so $g(x)\gt0$ there.
6 $f'(x)=\dfrac{1-\log x}{x^2}\lt0$ for $x\gt e$ (where $\log x\gt1$), so $f$ is strictly decreasing on $[e,\infty)$. Since $e\lt3\lt4$: $\dfrac{\log3}3\gt\dfrac{\log4}4$, i.e. $\log3^{1/3}\gt\log4^{1/4}$, and $\log$ is strictly increasing, so $3^{1/3}\gt4^{1/4}$. (Values: $1.4422\gt1.4142$.)
7 $f'=4\sin^3x\cos x-4\cos^3x\sin x$ $=-4\sin x\cos x(\cos^2x-\sin^2x)$ $=-2\sin2x\cos2x$ $=-\sin4x$. On $\left(0,\tfrac\pi4\right)$, $\sin4x\gt0$ so $f'\lt0$; on $\left(\tfrac\pi4,\tfrac\pi2\right)$, $\sin4x\lt0$ so $f'\gt0$. Decreasing on $\left[0,\tfrac\pi4\right]$, increasing on $\left[\tfrac\pi4,\tfrac\pi2\right]$.
IV
Applications.
  • 1 Wildlife. The population of a deer herd in a reserve $t$ years after a survey is modelled by $P(t)=500+60t-3t^2$ for $0\le t\le20$. When is the herd growing and when is it shrinking?
  • 2 Business. A firm's daily profit from $x$ items is $P(x)=-0.5x^2+60x-400$ rupees. For which outputs does profit increase with output?
  • 3 Medicine. The concentration of a painkiller in the blood $t$ hours after a tablet is taken is $C(t)=50te^{-0.5t}$ mg/L. Over which time interval is the concentration rising?
  • 4 Weather. The temperature in a city $t$ hours after 9 a.m. on a summer day is modelled by $T(t)=30+6\sin\left(\dfrac{\pi t}{12}\right)$ °C for $0\le t\le24$. When is the temperature rising?
1 $P'(t)=60-6t$: positive for $t\lt10$, negative for $t\gt10$. Growing on $[0,10]$ years, shrinking on $[10,20]$.
2 $P'(x)=-x+60\gt0$ for $x\lt60$: profit increases for $0\le x\le60$ items, decreases beyond.
3 $C'(t)=50e^{-0.5t}(1-0.5t)$, positive for $t\lt2$: rising for the first $2$ hours, falling after.
4 $T'(t)=\dfrac{\pi}{2}\cos\left(\dfrac{\pi t}{12}\right)\gt0$ when $\dfrac{\pi t}{12}\in\left[0,\tfrac\pi2\right)\cup\left(\tfrac{3\pi}2,2\pi\right]$, i.e. $0\le t\lt6$ or $18\lt t\le24$. Rising from 9 a.m. to 3 p.m. and from 3 a.m. to 9 a.m. the next day; falling from 3 p.m. to 3 a.m.
🎯 Exam corner Board · 1 markBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: almost every Board paper has a 3-mark "find the intervals in which $f$ is strictly increasing or decreasing" (usually a cubic or quartic, sometimes trigonometric on $[0,2\pi]$), plus 1-mark MCQs and Assertion–Reason items on functions like $x^3$. JEE asks for the parameter values that make a function monotonic and uses monotonicity to compare numbers.

E1. Board · 1 mark The function $f(x)=x^3-3x^2+3x$ is  (A) strictly increasing on $\mathbb R$  (B) strictly decreasing on $\mathbb R$  (C) neither increasing nor decreasing on $\mathbb R$  (D) strictly decreasing on $(0,1)$

Idea $f'(x)=3x^2-6x+3=3(x-1)^2\ge0$, and $f'=0$ only at $x=1$.
Answer (A). Trap (C)/(D): seeing $f'(1)=0$ and assuming the function must turn there. An isolated zero of $f'$ with no sign change does not stop $f$ from being strictly increasing.

E2. Board · 1 mark Assertion–Reason. Assertion (A): $f(x)=x^3$ is strictly increasing on $\mathbb R$. Reason (R): $f'(x)\gt0$ for every $x\in\mathbb R$.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea A is true: $x_1\lt x_2\Rightarrow x_1^3\lt x_2^3$ (or $f'=3x^2\ge0$ with an isolated zero). R is false: $f'(0)=0$.
Answer (C). The converse of Theorem 6.2(1) fails — the most tested subtlety of this section.

E3. Board · 3 marks Find the intervals in which $f(x)=3x^4-4x^3-12x^2+5$ is (a) strictly increasing, (b) strictly decreasing.

Derivative $f'(x)=12x^3-12x^2-24x$ $=12x(x^2-x-2)$ $=12x(x+1)(x-2)$; zeros $-1,0,2$. (1 mark)
Signs $f'(-2)=12(-2)(-1)(-4)=-96\lt0$; $f'\left(-\tfrac12\right)=12\left(-\tfrac12\right)\left(\tfrac12\right)\left(-\tfrac52\right)=7.5\gt0$; $f'(1)=12(1)(2)(-1)=-24\lt0$; $f'(3)=12(3)(4)(1)=144\gt0$. (1 mark)
Answer (a) Strictly increasing on $[-1,0]$ and $[2,\infty)$; (b) strictly decreasing on $(-\infty,-1]$ and $[0,2]$. (1 mark)

E4. Board · Case study A start-up's monthly sales (in hundreds of units) $t$ months after launch are modelled by $S(t)=t^3-15t^2+63t+50$, $0\le t\le12$. (i) Find $S'(t)$. [1 mark] (ii) Find the intervals in $[0,12]$ on which sales are increasing and decreasing. [2 marks] (iii) Are sales rising or falling at $t=5$, and how fast? [1 mark]

(i) $S'(t)=3t^2-30t+63$ $=3(t^2-10t+21)$ $=3(t-3)(t-7)$.
(ii) $S'\gt0$ on $[0,3)$ and $(7,12]$, $S'\lt0$ on $(3,7)$. Sales increase on $[0,3]$ and $[7,12]$ and decrease on $[3,7]$.
(iii) $S'(5)=3(2)(-2)=-12$: falling, at $12$ hundred units per month.

E5. JEE Main The function $f(x)=\sin x-ax$ is strictly decreasing on $\mathbb R$ if and only if  (A) $a\gt1$  (B) $a\ge1$  (C) $a\le1$  (D) $a\lt-1$

Method $f'(x)=\cos x-a$. We need $f'\le0$ everywhere with only isolated zeros, i.e. $a\ge\cos x$ for all $x$: $a\ge1$. At $a=1$, $f'=\cos x-1$ vanishes only at $x=2n\pi$, so $f$ is still strictly decreasing. For $a\lt1$, $f'\gt0$ near $x=0$.
Answer (B). Trap (A) wrongly excludes $a=1$ by demanding $f'\lt0$ everywhere. ⏱ Fast route: $a$ must beat the largest value of $\cos x$, which is $1$; equality is allowed because the zeros are isolated.

E6. JEE Advanced (One or more options correct.) Let $f(x)=\dfrac{\log x}{x}$, $x\gt0$. Then  (A) $f$ is strictly increasing on $(0,e]$  (B) $f$ is strictly decreasing on $[e,\infty)$  (C) $e^\pi\gt\pi^e$  (D) $3^{1/3}\lt2^{1/2}$

(A), (B) $f'(x)=\dfrac{1-\log x}{x^2}$: positive for $0\lt x\lt e$, negative for $x\gt e$. Both true.
(C) $\pi\gt e$, so by (B) $f(\pi)\lt f(e)$: $\dfrac{\log\pi}{\pi}\lt\dfrac1e$, i.e. $e\log\pi\lt\pi$, so $\pi^e\lt e^\pi$. True.
(D) $2\lt e\lt3$ lie on opposite sides of $e$, so monotonicity alone does not decide. Directly: $3^{1/3}\approx1.4422$ and $2^{1/2}\approx1.4142$ (equivalently $3^2=9\gt2^3=8$). So $3^{1/3}\gt2^{1/2}$ and (D) is false.
Answer (A), (B), (C). ⏱ Fast route for (D): raise both to the 6th power — compare $9$ and $8$.
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. During which months does a reservoir near you fill, and during which does it drain? Fit a smooth model to real storage data and use the sign of its derivative to find the intervals of increase and decrease.

Why it matters. Water boards decide when to release water for irrigation and drinking by watching whether storage is rising or falling — and how fast. India's Central Water Commission publishes weekly storage levels of major reservoirs, and similar open data exist in most countries.

What to do.

  1. Download (or copy from a weekly bulletin) the live storage of one reservoir for $12$ months — data.gov.in or the Central Water Commission website are good sources. Record month number $t$ and storage $S$.
  2. Plot $S$ against $t$ in a spreadsheet and fit a smooth trend: a polynomial of degree $3$, or a model $S(t)=a+b\sin\left(\dfrac{\pi t}{6}\right)+c\cos\left(\dfrac{\pi t}{6}\right)$ using least squares (the spreadsheet's LINEST, with columns $\sin$ and $\cos$).
  3. Differentiate your model by hand, solve $S'(t)=0$, and make a sign chart. State the intervals of increase and decrease, and compare with the monsoon calendar.
  4. Check: compute the month-to-month changes in the raw data. Do their signs agree with your sign chart?

Tools: Spreadsheet with charting and LINEST (or any curve-fitting app); open reservoir data.

Hand in: The data table, your fitted model and its derivative, the sign chart, and a paragraph explaining in plain words when the reservoir fills and drains — and one reason your model might disagree with the data.

Going further. Repeat with daily maximum temperatures for your city, or with Google Trends data for a seasonal search term such as "umbrella".

Instructor note: for $S=a+b\sin(\pi t/6)+c\cos(\pi t/6)$, $S'=\tfrac\pi6\left(b\cos\tfrac{\pi t}6-c\sin\tfrac{\pi t}6\right)$, zero where $\tan\tfrac{\pi t}6=\tfrac bc$ — two months six months apart, as expected for one fill-and-drain cycle per year. Real storage curves are steeper while filling than while draining, so a degree-3 polynomial often fits a single water year better.

🧪 Self-check quiz

0/5 correct

Q1. $f$ is continuous on $[a,b]$ and $f'(x)\gt0$ for all $x\in(a,b)$. Then on $[a,b]$, $f$ is

Q2. $f(x)=x^2-6x+5$ is strictly decreasing on

Q3. Which statement about $f(x)=x^3$ is correct?

Q4. $f(x)=e^{-x}$ is

Q5. $f(x)=x+\dfrac1x$ is strictly decreasing on

← 6.1 Rate of Change of Quantities6.3 Maxima and Minima: Local Extrema and Derivative Tests →