Maxima & Minima

Q: What is the difference between local and global maximum?

A local maximum is the largest value of f in a small neighbourhood of a point; a global maximum is the largest value on the whole domain or interval and must actually be attained. A function can have many local maxima but at most one global maximum value.

Q: What are critical points of a function?

Critical points are interior points of the domain where f'(x) = 0 or where f'(x) does not exist. Every local extremum inside an interval occurs at a critical point, so they are the only candidates to test, along with the end points for global questions.

Q: What is the first derivative test for maxima and minima?

At a critical point c, look at the sign of f' just before and just after c. A change from positive to negative gives a local maximum, from negative to positive a local minimum, and no change means there is no extremum at c.

Q: When does the second derivative test fail?

It fails when f''(c) = 0 (and of course where f'' does not exist). Then use the first derivative test or the first non-zero higher derivative: an even-order derivative gives an extremum, an odd-order one gives none. For example x⁴ has a minimum at 0 although f''(0) = 0.

Q: How do you find the absolute maximum and minimum on a closed interval?

Find the critical points inside [a, b], evaluate f at these points and at a and b, and compare the numbers. The largest is the absolute maximum and the smallest the absolute minimum; no classification of the critical points is needed.

Q: Can a function have a maximum at a point where it is not differentiable?

Yes. Extrema only compare values: -|x| has a maximum at 0 where it has a corner, and a function can even have a maximum at a point of discontinuity. Such points are critical points because the derivative does not exist there.

MathsApplication Of DerivativesFor JEE aspirants

Maxima and minima are the highest and lowest values of a function: local ones compare a point only with its neighbours, global ones compare it with the whole domain. Derivatives find them quickly: extrema sit at critical points, where $f^{'} (x) = 0$ or $f^{'}$ does not exist, and the sign of $f^{'}$ or $f^{''}$ tells a maximum from a minimum. Maxima and minima appear in every JEE Main and JEE Advanced paper, from curve tests to optimisation of areas, volumes and distances.

On this page1Definitions2

f^{'} = 0

and sign change3Critical points4Global extrema5

f^{''}

and

n

th tests6Optimisation7Examples

Key Formulas - Quick Reference

★ Must learnLocal maximum at $c$ : $f (c) \geq f (x)$ for all $x$ in some $(c - h, c + h)$ ; local minimum: $f (c) \leq f (x)$ there
Global maximum on $E$ : $f (c) \geq f (x)$ for all $x \in E$ (the largest value, actually attained)
★ Must learnNecessary condition: $f$ differentiable at an extremum $c$ $\Rightarrow$ $f^{'} (c) = 0$ (not sufficient: $x^{3}$ at $0$ )
Critical points: interior points where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist; stationary points $\subseteq$ critical points
★ Must learnFirst derivative test: $f^{'}$ changes $+ \to -$ : maximum; $- \to +$ : minimum; no change: neither
★ Must learnSecond derivative test: $f^{'} (c) = 0$ and $f^{''} (c) < 0$ : maximum; $f^{''} (c) > 0$ : minimum; $f^{''} (c) = 0$ : test further
$n$ th derivative test: first non-zero derivative $f^{(n)} (c)$ ; $n$ even: extremum (max if $< 0$ , min if $> 0$ ); $n$ odd: none
★ Must learnOn $[a, b]$ : global max/min $=$ largest/smallest of $f (a)$ , $f (b)$ and $f$ at the critical points

1. Local and Global Extrema

Global maximum. $f$ has a global maximum on a set $E$ if there is $c \in E$ with $f (x) \leq f (c)$ for all $x \in E$ . The value $f (c)$ is the global (absolute) maximum. Global minimum: $f (x) \geq f (c)$ for all $x \in E$ .

Local maximum. $f$ has a local maximum at $x = c$ if $f (c)$ is the greatest value of $f$ in some small neighbourhood $(c - h, c + h)$ : $f (c - δ) \leq f (c) \geq f (c + δ)$ for $0 < δ < h$ . Local minimum: $f (c - δ) \geq f (c) \leq f (c + δ)$ .

A maximum or a minimum is called an extremum. If $c$ is an end point of the domain, use only the side that exists: $(c - h, c]$ or $[c, c + h)$ .

Figure 1: Local maxima (green) at

x = a, c_{2}, c_{4}

and local minima (red) at

x = c_{1}, c_{3}, b

. The largest of all values,

f (c_{2})

, is the global maximum; the smallest,

f (c_{3})

, is the global minimum. End points can be local extrema (compare with one side only), and an extremum can sit at a corner (

c_{3}

) where

f^{'}

does not exist.

Local maximum

Largest value in a small neighbourhood. A function can have many; a local maximum can be smaller than a local minimum elsewhere.

Global maximum

Largest value on the whole set, and it must be attained. At most one value (possibly at several points); may not exist.

Extrema do not need a derivative, or even continuity. Only the comparison of $f (c)$ with nearby values matters.

Figure 2: Extrema need not involve a derivative. Panel 1:

f (x) = ∣ x ∣

for

0 < ∣ x ∣ \leq 2

with

f (0) = 1

; nearby values are smaller than

1

, so

x = 0

is a local maximum (Solved Example 1). Panels (i) to (iii) are for Solved Example 25: compare

f (a)

with the values close to

a

on both sides.

2. Extrema of Differentiable Functions

2.1 A necessary condition

Theorem. If $f$ has an extremum at $x = c$ and $f$ is differentiable at $c$ , then $f^{'} (c) = 0$ .

The converse is false: $f^{'} (c) = 0$ does not guarantee an extremum. For $f (x) = x^{3}$ , $f^{'} (0) = 0$ but $f (0) = 0$ is neither a maximum nor a minimum, because $f$ keeps increasing through $0$ .

2.2 A sufficient condition: the sign change

Theorem. For a differentiable $f$ , $f (c)$ is an extremum if and only if $f^{'}$ changes sign as $x$ passes through $c$ : from positive to negative gives a local maximum, from negative to positive a local minimum.

Points where $f^{'} (x) = 0$ are called stationary points: the rate of change of $f$ is zero there and the tangent is horizontal.

Figure 3: First derivative test at a stationary point

c

(

f^{'} (c) = 0

, horizontal tangent). Left: tangents tilt up before

c

and down after it, so

f (c)

is a local maximum. Middle: down then up, a local minimum. Right:

f (x) = (x - 2)^{3}

-type behaviour, up on both sides: no extremum even though

f^{'} (c) = 0

2.3 First derivative test

Find $f^{'} (x)$ (for $f$ continuous and differentiable).
Solve $f^{'} (x) = 0$ : the stationary points $c$ .
Observe the sign of $f^{'}$ as $x$ crosses each $c$ from left to right: $- \to +$ minimum, $+ \to -$ maximum, no change: neither.

Figure 4:

f (x) = x^{3} - 12 x

f^{'} (x) = 3 (x - 2) (x + 2)

. The sign chart reads

+, -, +

: a local maximum

f (- 2) = 16

and a local minimum

f (2) = - 16

. For a continuous function, maxima and minima alternate.

For a continuous function, maxima and minima alternate along the $x$ -axis: between two local maxima there is a local minimum, and the other way round.

Exam Trick

Cubic: maximum at the smaller root. If $f^{'} (x) = 3 (x - α) (x - β)$ with $α < β$ (positive leading coefficient), the signs are $+, -, +$ : maximum at $α$ , minimum at $β$ . "Positive point of maximum" means $α > 0$ , i.e. both roots positive and distinct: $D > 0$ , sum of roots $> 0$ , $f^{'} (0) > 0$ .

Key idea

A smooth extremum needs

f^{'} (c) = 0

, but it is the sign change of

f^{'}

that makes it a maximum or a minimum.

3. Critical Points and Continuous Functions

For a continuous function $f$ , the interior points of the domain where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist are the critical points. Every stationary point is a critical point, but not conversely.

Important: for $f$ defined on a subset of $R$ , interior extrema (if any) occur only at critical points. Critical points are always interior points of an interval; end points are checked separately.

Figure 5:

f (x) = max (sin x, cos x)

(solid) on

(0, 2 π)

. It has corners where

sin x = cos x

, at

x = \frac{π}{4}

and

\frac{5 π}{4}

(

f^{'}

does not exist), and a smooth top at

x = \frac{π}{2}

(

f^{'} = 0

). These three are the critical points.

Examples: $f (x) = ∣ x^{2} - 2 x ∣$ has critical points $x = 0, 2$ (corners, $f^{'}$ does not exist) and $x = 1$ ( $f^{'} (1) = 0$ ). For a piecewise function such as $x^{3} + x^{2} - 10 x$ ( $x < 0$ ), $3 sin x$ ( $x \geq 0$ ), the joining point $x = 0$ is critical because the one-sided derivatives $- 10$ and $3$ differ.

Stationary point

$f^{'} (c) = 0$ . The tangent is horizontal. Example: $x^{2}$ at $0$ .

Critical point

$f^{'} (c) = 0$ or $f^{'} (c)$ does not exist (interior). Example: $∣ x ∣$ at $0$ (no tangent slope).

Quick Recall: tap to check

Is every stationary point an extremum?

No.

x^{3}

has a stationary point at

0

but no extremum.

What are the critical points of

∣ x^{2} - 2 x ∣

x = 0, 2

(derivative does not exist) and

x = 1

(derivative zero).

f^{'} (x) = 4 x (x - 1) (x - 2)

, where are the local maxima and minima?

Signs

-, +, -, +

: minima at

0

and

2

, maximum at

1

4. Global Extrema of Continuous Functions

4.1 On a closed interval $[a, b]$

Find the critical points $c_{1}, c_{2}, \dots, c_{n}$ in $(a, b)$ .
Compute $f (a), f (c_{1}), \dots, f (c_{n}), f (b)$ .
$M = max {f (a), f (c_{1}), \dots, f (c_{n}), f (b)}$ is the global maximum and $m = min {\dots}$ the global minimum. (A continuous function on a closed interval always has both.)

Figure 6: On

[- 1, 3]

the candidates for

f (x) = x^{3} - 12 x

are the critical point

x = 2

and the end points:

f (- 1) = 11

f (2) = - 16

f (3) = - 9

. Greatest value

11

, least value

- 16

. The local maximum at

x = - 2

lies outside the interval and is ignored.

Exam Trick

No tests needed on $[a, b]$ . For greatest and least values on a closed interval, do not classify the critical points at all: just evaluate $f$ at the critical points inside and at the two ends, and pick the largest and smallest numbers.

4.2 On an open interval $(a, b)$

Find the critical points $c_{1}, \dots, c_{n}$ and the values $f (c_{1}), \dots, f (c_{n})$ ; let $M$ be their maximum and $m$ their minimum.
Find the end limits $ℓ_{1} = x \to a^{+} lim f (x)$ and $ℓ_{2} = x \to b^{-} lim f (x)$ ; let $ℓ = min {ℓ_{1}, ℓ_{2}}$ and $L = max {ℓ_{1}, ℓ_{2}}$ .
If $m \leq ℓ$ , $m$ is the global minimum; if $m > ℓ$ , there is no global minimum. If $M \geq L$ , $M$ is the global maximum; if $M < L$ , there is no global maximum.

A value that is only approached, never attained, cannot be a global extremum. The graphs below show how open ends and jumps affect the answer.

Figure 7: Graphs for Solved Example 26. Filled dots are values taken by the function; hollow dots are values only approached. A global extremum must be attained, so an extreme value at a hollow dot does not count.

Figure 8: Solved Example 14. On

(0, \frac{3}{2})

the values approach

1

and

- 1

without reaching them, so they decide nothing unless

f (0) = λ

reaches one of them. With

λ = 0

there is no global maximum or minimum; with

λ = 1

the global maximum

1

is attained at

x = 0

. A global maximum exists exactly when

λ \geq 1

Key idea

Closed interval: compare critical values with end values. Open interval: also compare with the end limits, which do not count unless attained.

5. Second and Higher Derivative Tests

5.1 Second derivative test

Find $f^{'} (x)$ and solve $f^{'} (x) = 0$ ; let $x = c$ be a solution.
Find $f^{''} (c)$ .
If $f^{''} (c) < 0$ : local maximum. If $f^{''} (c) > 0$ : local minimum. If $f^{''} (c) = 0$ : the test gives no answer; investigate further.

Reason: at a maximum $f^{'}$ changes from positive to negative, so $f^{'}$ is decreasing there and $f^{''} (c) \leq 0$ ; similarly $f^{''} (c) \geq 0$ at a minimum.

5.2 The $n$ th derivative test

Suppose $f^{'} (c) = f^{''} (c) = \dots = f^{(n - 1)} (c) = 0$ and $f^{(n)} (c) \neq = 0$ .

$n$	Sign of $f^{(n)} (c)$	Conclusion at $x = c$
even	$< 0$	local maximum
even	$> 0$	local minimum
odd	$< 0$	no extremum; $f$ decreasing at $c$
odd	$> 0$	no extremum; $f$ increasing at $c$

Figure 9: Left:

f (x) = sin 2 x - x

has

f^{'} = 2 cos 2 x - 1 = 0

\frac{π}{6}, \frac{5 π}{6}

;

f^{''} = - 4 sin 2 x

is negative at

\frac{π}{6}

(maximum) and positive at

\frac{5 π}{6}

(minimum). Right:

x^{5} - 5 x^{4} + 5 x^{3} - 1

has a maximum at

1

and a minimum at

3

; at

0

f^{''} = 0

but

f^{'''} = 30 \neq = 0

, an odd order, so there is no extremum: the graph flattens and keeps rising.

When $f^{''} (c) = 0$ , the first derivative test is usually quicker than computing higher derivatives: for $x^{5} - 5 x^{4} + 5 x^{3} - 1$ , $f^{'} (x) = 5 x^{2} (x - 1) (x - 3)$ has the same sign on both sides of $0$ (the factor $x^{2}$ ), so there is no extremum at $0$ .

5.3 Extrema of related functions

If $g = \frac{k}{f}$ with $k > 0$ , then $g^{'} = - \frac{k f ^{'}}{f ^{2}}$ : $g$ has the same critical points as $f$ , with maxima and minima interchanged, and vertical asymptotes at the zeros of $f$ .

Figure 10: Solved Example 15.

f

has local minima

f (- 3) = - 75

f (1) = 53

and a local maximum

f (0) = 60

; its zeros

α \approx - 3.87

and

β \approx - 1.61

. For

g = \frac{40}{f}

g^{'} = - \frac{40 f ^{'}}{f ^{2}}

: the same critical points with the roles swapped (local maxima at

- 3

and

1

, local minimum at

0

), vertical asymptotes at

α

and

β

, and

y = 0

x \to \pm \infty

JEE Advanced

Composition rule. If $f$ has a local maximum at $c$ and $h$ is strictly increasing on the range of $f$ , then $h \circ f$ also has a local maximum at $c$ ; if $h$ is strictly decreasing, it becomes a local minimum. So $e^{f}$ , $ln f$ (for $f > 0$ ), $f$ and $f^{3}$ keep the type, while $\frac{1}{f}$ (on an interval where $f$ keeps one sign) and $- f$ reverse it. This is why one may maximise $A^{2}$ instead of $A$ , or $ln y$ instead of $y = (\frac{1}{x})^{x}$ : same critical points, simpler algebra.

Quick Recall: tap to check

f^{'} (c) = 0

and

f^{''} (c) = 0

. Is there an extremum at

c

Not decided. Use the sign change of

f^{'}

, or the first non-zero higher derivative (

x^{4}

0

is a minimum,

x^{3}

is not).

f^{'} (c) = f^{''} (c) = 0

f^{'''} (c) = 6

. Conclusion?

Odd order: no extremum;

f

is increasing at

c

How are the extrema of

\frac{1}{f}

related to those of

f

Same critical points (where

f \neq = 0

), with maxima and minima swapped.

Key idea

f^{''} (c) < 0

maximum,

f^{''} (c) > 0

minimum; if

f^{''} (c) = 0

, go back to the sign of

f^{'}

or to the first non-zero higher derivative.

6. Applications: Optimisation

In an applied problem we build an objective function (area, volume, cost, distance) in terms of one variable and find its extreme value.

Draw a figure and name the quantities. Write the quantity to be optimised.
Use the given condition (fixed perimeter, fixed volume, a point on a curve, similar triangles) to express it in one variable.
Write the domain of that variable.
Solve $f^{'} (x) = 0$ in the domain and confirm the type (sign change or $f^{''}$ ); for a global answer, compare with the end values too.

Solid / figure	Volume or area	Surface area
Cuboid $l \times b \times h$	$V = l bh$	$S = 2 (l b + bh + h l)$
Cube of edge $a$	$V = a^{3}$	$S = 6 a^{2}$
Right circular cone	$V = \frac{1}{3} π r^{2} h$	curved $= π r l$ ( $l$ = slant height)
Right circular cylinder	$V = π r^{2} h$	curved $= 2 π r h$ ; total $= 2 π r h + 2 π r^{2}$
Sphere	$V = \frac{4}{3} π r^{3}$	$S = 4 π r^{2}$
Sector of a circle	area $= \frac{1}{2} r^{2} θ$ ( $θ$ in radians)	arc $= r θ$
Prism	$V = (base area) \times h$	lateral $= (base perimeter) \times h$ ; total $=$ lateral $+ 2 \times$ base
Pyramid	$V = \frac{1}{3} (base area) \times h$	curved $= \frac{1}{2} (base perimeter) \times (slant height)$

Figure 11: Two classic set-ups. Left: a rectangle in a semicircle of radius

r

with a corner at

(r cos θ, r sin θ)

has area

A = r^{2} sin 2 θ

, largest at

θ = \frac{π}{4}

x = r 2

y = \frac{r}{2}

A = r^{2}

. Right: by similar triangles a cylinder of radius

x

in a cone of radius

r

, height

h

has height

y = \frac{h}{r} (r - x)

; its volume is largest for

x = \frac{2 r}{3}

y = \frac{h}{3}

(drawn to scale), which is

\frac{4}{9}

of the cone's volume.

Exam Trick

Standard optimum shapes (use them to check answers):

Rectangle of greatest area in a circle: a square. With a fixed perimeter: also a square.
Rectangle in a semicircle of radius $r$ : $x = r 2$ along the diameter, height $\frac{r}{2}$ , area $r^{2}$ .
Cylinder in a cone: radius $\frac{2}{3} r$ , height $\frac{h}{3}$ , volume $\frac{4}{9}$ of the cone. Closed cylinder of given surface and largest volume: $h = 2 r$ .
Open box from a square sheet of side $a$ : cut squares of side $\frac{a}{6}$ . Open tank with square base and fixed surface: height $=$ half the side.
$x + y$ fixed: $x^{m} y^{n}$ is largest when $\frac{x}{m} = \frac{y}{n}$ .

6.1 Shortest paths: the reflection trick

To minimise $P A + P B$ for $P$ on a line, avoid writing a messy function of $P$ . If $A$ and $B$ are on opposite sides, the straight segment $A B$ is shortest. If they are on the same side, reflect one of them in the line.

Figure 12: Least value of

P A + P B

for

P

on a line. Left: if

A

and

B

are on the same side, reflect

A

A^{'}

; then

P A + P B = P A^{'} + P B \geq A^{'} B

, with equality where

A^{'} B

meets the line (the light-ray path). Right:

A (1, 2)

and

B (- 2, - 4)

lie on opposite sides of

y = x

, so

P A + P B \geq A B

directly, with equality where

A B

(

y = 2 x

) meets

y = x

P = (0, 0)

Key idea

Optimisation = one-variable objective + domain + critical points + a check; geometry (reflection, standard shapes) often shortcuts the calculus.

Figure 13: Finding extrema. Local: first derivative test (always works for continuous

f

) or the second derivative test. Global on

[a, b]

: compare the values at critical points and end points. On an open interval, compare with the end limits.

Figure 14: Mind map of maxima and minima.

7. Solved Examples

Solved Example 1

Let

f (x) = ∣ x ∣

for

0 < ∣ x ∣ \leq 2

and

f (0) = 1

. Examine the behaviour of

f

x = 0

Solution:

Near $x = 0$ , $f (x) = ∣ x ∣$ is small (less than $1$ for $0 < ∣ x ∣ < 1$ ), while $f (0) = 1$ . So $f (0) > f (x)$ for all $x$ in $(- 1, 1)$ , $x \neq = 0$ (Figure 2, panel 1).

Answer: $f$ has a local maximum at $x = 0$ , although it is not even continuous there.

Solved Example 2

Let

f (x) = - x^{3} + \frac{b ^{3} - b ^{2} + b - 1}{b ^{2} + 3 b + 2}

for

0 \leq x < 1

and

f (x) = 2 x - 3

for

1 \leq x \leq 3

. Find all possible values of

b

such that

f

has its smallest value at

x = 1

Solution:

On $[1, 3]$ , $2 x - 3$ increases from $f (1) = - 1$ . On $[0, 1)$ , $- x^{3} + k$ decreases (with $k$ the constant), so its values stay above its limit at $x \to 1^{-}$ .
The smallest value is at $x = 1$ exactly when $x \to 1^{-} lim f (x) \geq f (1)$ : $- 1 + \frac{b ^{3} - b ^{2} + b - 1}{b ^{2} + 3 b + 2} \geq - 1$ .
$\frac{( b - 1 ) ( b ^{2} + 1 )}{( b + 1 ) ( b + 2 )} \geq 0$ ; since $b^{2} + 1 > 0$ , this needs $\frac{b - 1}{( b + 1 ) ( b + 2 )} \geq 0$ .

Answer: $b \in (- 2, - 1) \cup [1, \infty)$ .

Solved Example 3

Find the stationary points of

f (x) = 4 x^{3} - 6 x^{2} - 24 x + 9

Solution:

$f^{'} (x) = 12 x^{2} - 12 x - 24 = 12 (x - 2) (x + 1) = 0$ gives $x = - 1, 2$ .
$f (- 1) = - 4 - 6 + 24 + 9 = 23$ ; $f (2) = 32 - 24 - 48 + 9 = - 31$ .

Answer: $(- 1, 23)$ and $(2, - 31)$ .

Solved Example 4

f (x) = x^{3} + a x^{2} + b x + c

has extreme values at

x = - 1

and

x = 3

, find

a

b

c

Solution:

$f$ is differentiable, so $f^{'} (- 1) = f^{'} (3) = 0$ , where $f^{'} (x) = 3 x^{2} + 2 a x + b$ .
$f^{'} (3) = 27 + 6 a + b = 0$ and $f^{'} (- 1) = 3 - 2 a + b = 0$ . Subtracting: $24 + 8 a = 0$ .

Answer: $a = - 3$ , $b = - 9$ , $c$ any real number.

Solved Example 5

Find the points of maxima and minima of

f (x) = x^{2} (x - 2)^{2}

Solution:

$f^{'} (x) = 4 x (x - 1) (x - 2) = 0$ at $x = 0, 1, 2$ .
Sign of $f^{'}$ : $-, +, -, +$ across $0, 1, 2$ .

Answer: maximum at $x = 1$ ; minima at $x = 0$ and $x = 2$ .

Solved Example 6

Find the points of maxima and minima of

f (x) = x^{3} - 12 x

and draw the graph.

Solution:

$f^{'} (x) = 3 (x^{2} - 4) = 3 (x - 2) (x + 2)$ ; signs $+, -, +$ across $- 2, 2$ .
Maximum at $x = - 2$ : $f (- 2) = 16$ ; minimum at $x = 2$ : $f (2) = - 16$ . Roots: $0, \pm 23$ .

Answer: local maximum $16$ at $x = - 2$ , local minimum $- 16$ at $x = 2$ (Figure 4).

Solved Example 7

Show that

f (x) = x^{3} - 6 x^{2} + 12 x - 8

has no point of local maximum or minimum. Hence draw its graph.

Solution:

$f^{'} (x) = 3 (x^{2} - 4 x + 4) = 3 (x - 2)^{2} = 0$ only at $x = 2$ .
$f^{'} (x) > 0$ on both sides of $2$ : no sign change.

Answer: no extremum; $f (x) = (x - 2)^{3}$ is increasing on $R$ with a horizontal tangent (and an inflection) at $(2, 0)$ , as in Figure 3, right panel.

Solved Example 8

Let

f (x) = x^{3} + 3 (a - 7) x^{2} + 3 (a^{2} - 9) x - 1

. If

f

has a positive point of maximum, find the possible values of

a

Solution:

$f^{'} (x) = 3 [x^{2} + 2 (a - 7) x + (a^{2} - 9)]$ . Let $α < β$ be its roots; signs $+, -, +$ , so the maximum is at $α$ .
$α > 0$ means both roots positive and distinct: (i) $D > 0$ : $(a - 7)^{2} - (a^{2} - 9) = 58 - 14 a > 0$ , so $a < \frac{29}{7}$ .
(ii) Sum of roots $= 2 (7 - a) > 0$ : $a < 7$ . (iii) Product $= a^{2} - 9 > 0$ (that is, $f^{'} (0) > 0$ ): $a < - 3$ or $a > 3$ .

Answer: $a \in (- \infty, - 3) \cup (3, \frac{29}{7})$ .

Solved Example 9

Find the critical points of

f (x) = max (sin x, cos x)

x \in (0, 2 π)

Solution:

$f = cos x$ on $(0, \frac{π}{4}]$ , $sin x$ on $[\frac{π}{4}, \frac{5 π}{4}]$ , $cos x$ on $[\frac{5 π}{4}, 2 π)$ . At $\frac{π}{4}$ and $\frac{5 π}{4}$ the graph has corners ( $f^{'}$ does not exist); on the middle piece $f^{'} = cos x = 0$ at $\frac{π}{2}$ (Figure 5).

Answer: $x = \frac{π}{4}, \frac{π}{2}, \frac{5 π}{4}$ .

Solved Example 10

Find the possible points of maxima or minima of

f (x) = ∣ x^{2} - 2 x ∣

x \in R

Solution:

$f (x) = x^{2} - 2 x$ for $x \leq 0$ or $x \geq 2$ , and $2 x - x^{2}$ for $0 < x < 2$ .
$f^{'} (x) = 2 (x - 1)$ for $x < 0$ or $x > 2$ , and $2 (1 - x)$ for $0 < x < 2$ .
$f^{'} (1) = 0$ ; at $x = 0$ and $x = 2$ the one-sided derivatives differ ( $\mp 2$ ), so $f^{'}$ does not exist.

Answer: the critical points $x = 0, 1, 2$ (minima $0$ at $x = 0, 2$ ; maximum $1$ at $x = 1$ ).

Solved Example 11

Let

f (x) = x^{3} + x^{2} - 10 x

for

x < 0

and

f (x) = 3 sin x

for

x \geq 0

. Examine the behaviour of

f

x = 0

Solution:

$f$ is continuous at $0$ (both pieces give $0$ ).
$f^{'} (0^{-}) = 3 (0) + 2 (0) - 10 = - 10$ and $f^{'} (0^{+}) = 3 cos 0 = 3$ : $f$ is not differentiable at $0$ , so $x = 0$ is a critical point.
$f^{'}$ changes from negative to positive across $0$ .

Answer: local minimum at $x = 0$ .

Solved Example 12

Find the critical points of

f (x) = 4 x^{3} - 6 x^{2} - 24 x + 9

if (i)

x \in [0, 3]

(ii)

x \in [- 3, 3]

(iii)

x \in [- 1, 2]

Solution:

$f^{'} (x) = 12 (x - 2) (x + 1) = 0$ at $x = - 1, 2$ . Critical points must be interior points.

(i) $[0, 3]$ : only $x = 2$ .
(ii) $[- 3, 3]$ : $x = - 1$ and $x = 2$ .
(iii) $[- 1, 2]$ : none, because $x = - 1$ and $x = 2$ are end points.

Answer: (i) $2$ (ii) $- 1, 2$ (iii) no critical point.

Solved Example 13

Find the greatest and least values of

f (x) = x^{3} - 12 x

x \in [- 1, 3]

Solution:

$f^{'} (x) = 3 (x - 2) (x + 2)$ : the only critical point in $(- 1, 3)$ is $x = 2$ .
$f (- 1) = 11$ , $f (2) = - 16$ , $f (3) = - 9$ .

Answer: greatest value $11$ (at $x = - 1$ ), least value $- 16$ (at $x = 2$ ) (Figure 6).

Solved Example 14

Let

f (x) = x^{2} + x

for

- 1 \leq x < 0

f (0) = λ

, and

f (x) = lo g_{1/2} (x + \frac{1}{2})

for

0 < x < \frac{3}{2}

. Discuss the global maxima and minima for

λ = 0

and

λ = 1

. For which

λ

does

f

have a global maximum?

Solution:

On $[- 1, 0)$ : values from $0$ (at $x = - 1$ ) down to $- \frac{1}{4}$ (at $x = - \frac{1}{2}$ ), tending to $0$ as $x \to 0^{-}$ .
On $(0, \frac{3}{2})$ : the logarithm decreases from $x \to 0^{+} lim f (x) = 1$ to $x \to 3/ 2^{-} lim f (x) = lo g_{1/2} 2 = - 1$ ; neither limit is attained.
$λ = 0$ : values come arbitrarily close to $1$ and to $- 1$ without reaching them: no global maximum, no global minimum.
$λ = 1$ : the value $1$ is attained at $x = 0$ : global maximum $1$ ; still no global minimum.

Answer: a global maximum exists exactly when $f (0) = λ \geq 1$ (Figure 8).

Solved Example 15

Find the extrema of

f (x) = 3 x^{4} + 8 x^{3} - 18 x^{2} + 60

. Draw the graph of

g (x) = \frac{40}{f ( x )}

and comment on its local and global extrema.

Solution:

$f^{'} (x) = 12 x (x^{2} + 2 x - 3) = 12 x (x + 3) (x - 1)$ ; signs $-, +, -, +$ across $- 3, 0, 1$ .
Local minima $f (- 3) = - 75$ , $f (1) = 53$ ; local maximum $f (0) = 60$ . $f \to \infty$ at both ends, so no global maximum; global minimum $- 75$ .
$g^{'} (x) = - \frac{40}{f ^{2}} f^{'} (x)$ : same critical points, opposite signs. So $g$ has local maxima at $x = - 3$ ( $- \frac{8}{15}$ ) and $x = 1$ ( $\frac{40}{53}$ ) and a local minimum at $x = 0$ ( $\frac{2}{3}$ ).
$f$ has two real zeros $α \approx - 3.87$ , $β \approx - 1.61$ : vertical asymptotes of $g$ ; $g \to \pm \infty$ near them and $g \to 0$ as $x \to \pm \infty$ .

Answer: as above; $g$ has no global extrema (Figure 10).

Solved Example 16

Find the points of local maxima or minima of

f (x) = sin 2 x - x

x \in (0, π)

Solution:

$f^{'} (x) = 2 cos 2 x - 1 = 0$ : $cos 2 x = \frac{1}{2}$ , so $x = \frac{π}{6}, \frac{5 π}{6}$ .
$f^{''} (x) = - 4 sin 2 x$ : $f^{''} (\frac{π}{6}) = - 23 < 0$ and $f^{''} (\frac{5 π}{6}) = 23 > 0$ .

Answer: maximum at $x = \frac{π}{6}$ , minimum at $x = \frac{5 π}{6}$ (Figure 9, left).

Solved Example 17

Find the points of local maxima or minima of

f (x) = x^{5} - 5 x^{4} + 5 x^{3} - 1

Solution:

$f^{'} (x) = 5 x^{2} (x - 1) (x - 3) = 0$ at $x = 0, 1, 3$ . $f^{''} (x) = 10 x (2 x^{2} - 6 x + 3)$ .
$f^{''} (1) = - 10 < 0$ : maximum at $1$ . $f^{''} (3) = 90 > 0$ : minimum at $3$ .
$f^{''} (0) = 0$ ; $f^{'''} (x) = 30 (2 x^{2} - 4 x + 1)$ , $f^{'''} (0) = 30 \neq = 0$ : odd order, so no extremum at $0$ . (Quicker: $f^{'}$ does not change sign at $0$ , signs $+, +, -, +$ .)

Answer: maximum at $x = 1$ , minimum at $x = 3$ , neither at $x = 0$ (Figure 9, right).

Solved Example 18

If the equation

x^{3} + p x + q = 0

has three distinct real roots, show that

4 p^{3} + 27 q^{2} < 0

Solution:

$f (x) = x^{3} + p x + q$ must have a local maximum above the axis and a local minimum below it. $f^{'} (x) = 3 x^{2} + p = 0$ needs $p < 0$ : $x = \pm s$ with $s = - \frac{p}{3}$ .
Maximum at $- s$ , minimum at $s$ : $f (- s) = q - \frac{2 p}{3} s$ , $f (s) = q + \frac{2 p}{3} s$ .
$f (- s) f (s) < 0$ : $q^{2} - \frac{4 p ^{2}}{9} s^{2} = q^{2} + \frac{4 p ^{3}}{27} < 0$ .

Answer: $4 p^{3} + 27 q^{2} < 0$ .

Solved Example 19

Find two positive numbers

x

and

y

such that

x + y = 60

and

x y^{3}

is maximum.

Solution:

$x = 60 - y$ , so $f (y) = (60 - y) y^{3}$ , $y \in (0, 60)$ .
$f^{'} (y) = 3 y^{2} (60 - y) - y^{3} = y^{2} (180 - 4 y) = 0$ gives $y = 45$ ; $f^{'}$ changes from $+$ to $-$ there.

Answer: $x = 15$ , $y = 45$ .

Solved Example 20

Rectangles are inscribed in a semicircle of radius

r

. Find the rectangle with maximum area.

Solution:

Let the side along the diameter be $x$ and the other side $y$ . A top corner lies on the circle: $\frac{x ^{2}}{4} + y^{2} = r^{2}$ , so $y = r^{2} - \frac{x ^{2}}{4}$ , with $x \in (0, 2 r)$ .
$A = x y$ ; maximise $A^{2} = r^{2} x^{2} - \frac{x ^{4}}{4}$ instead (same critical points): derivative $2 x (r^{2} - \frac{x ^{2}}{2}) = 0$ gives $x = r 2$ , a maximum.
Alternative: with the corner at $(r cos θ, r sin θ)$ , $A = 2 r cos θ \cdot r sin θ = r^{2} sin 2 θ$ , largest at $θ = \frac{π}{4}$ .

Answer: $x = r 2$ , $y = \frac{r}{2}$ , maximum area $r^{2}$ (Figure 11, left).

Solved Example 21

A sheet of area

40 m^{2}

is used to make an open tank with a square base. Find the dimensions of the base for which the volume of the tank is maximum.

Solution:

Base side $x$ , height $y$ : $x^{2} + 4 x y = 40$ , so $y = \frac{40 - x ^{2}}{4 x}$ , $x \in (0, 40)$ .
$V = x^{2} y = \frac{40 x - x ^{3}}{4}$ ; $V^{'} (x) = \frac{40 - 3 x ^{2}}{4} = 0$ gives $x = \frac{40}{3}$ .
$V^{''} (x) = - \frac{3 x}{2} < 0$ : maximum.

Answer: base side $\frac{40}{3} \approx 3.65 m$ (height $\approx 1.83 m$ , half the side; volume $\approx 24.3 m^{3}$ ).

Solved Example 22

A right circular cylinder is inscribed in a given cone of radius

r

and height

h

. Find the dimensions of the cylinder of maximum volume.

Solution:

Cylinder radius $x$ , height $y$ . Similar triangles: $\frac{y}{r - x} = \frac{h}{r}$ , so $y = \frac{h}{r} (r - x)$ , $x \in (0, r)$ .
$V (x) = π x^{2} y = \frac{π h}{r} (r x^{2} - x^{3})$ ; $V^{'} (x) = \frac{π h}{r} x (2 r - 3 x) = 0$ gives $x = \frac{2 r}{3}$ .
$V^{''} (\frac{2 r}{3}) = \frac{π h}{r} (2 r - 4 r) < 0$ : maximum.

Answer: radius $\frac{2 r}{3}$ , height $\frac{h}{3}$ (Figure 11, right).

Solved Example 23

Among all regular square pyramids of volume

362 cm^{3}

, find the dimensions of the pyramid with the least lateral surface area.

Solution:

Base side $x$ , height $y$ : $\frac{1}{3} x^{2} y = 362$ , so $y = \frac{108 2}{x ^{2}}$ .
Slant height of a face $l = \frac{x ^{2}}{4} + y^{2}$ ; lateral area $S = \frac{1}{2} (4 x) l = x^{4} + 4 x^{2} y^{2} = x^{4} + \frac{8 ( 108 ) ^{2}}{x ^{2}}$ .
Minimise $f (x) = x^{4} + \frac{8 ( 108 ) ^{2}}{x ^{2}}$ : $f^{'} (x) = 4 x^{3} - \frac{16 ( 108 ) ^{2}}{x ^{3}} = \frac{4 ( x ^{6} - 6 ^{6} )}{x ^{3}} = 0$ gives $x = 6$ , a minimum.

Answer: base side $6 cm$ , height $32 cm$ .

Solved Example 24

Let

A (1, 2)

and

B (- 2, - 4)

be fixed points. A variable point

P

is chosen on the line

y = x

so that the perimeter of

△ P A B

is minimum. Find

P

Solution:

$A B$ is fixed, so we minimise $P A + P B$ .
Check the sides of $y = x$ : for $A$ , $y - x = 1 > 0$ ; for $B$ , $y - x = - 2 < 0$ . They are on opposite sides.
Then $P A + P B \geq A B$ , with equality when $P$ lies on segment $A B$ . Line $A B$ : slope $\frac{- 4 - 2}{- 2 - 1} = 2$ , so $y = 2 x$ ; it meets $y = x$ at the origin.

Answer: $P = (0, 0)$ (Figure 12, right). If $A$ and $B$ were on the same side, reflect one of them in the line first (Figure 12, left).

Solved Example 25

In each graph of Figure 2, panels (i) to (iii), decide whether

x = a

is a point of local maximum, local minimum or neither.

Solution:

(i) $f (a)$ lies above the values on both sides: local maximum.
(ii) Values just left of $a$ are below $f (a)$ , values just right are above it: neither.
(iii) $f (a)$ lies below the values on both sides: local minimum.

Answer: (i) maximum (ii) neither (iii) minimum.

Solved Example 26

For each graph in Figure 7, identify the points of global maximum/minimum and local maximum/minimum.

Solution:

(i) On $(1, 3]$ : local maximum at $x = 2$ (value $3$ ), local minimum at the end $x = 3$ (value $2$ ). Global maximum $3$ at $x = 2$ ; no global minimum, since values approach $1$ without reaching it.
(ii) On $[- 1, 3)$ : local minimum at the end $x = - 1$ . At $x = 1$ , $f (1) = 3$ is below the values just to its left and above those to its right: neither. No global maximum ( $4$ not attained) and no global minimum ( $1$ not attained).
(iii) On $(- 1, 2)$ : local and global maximum at $x = 1$ (value $3$ ); local and global minimum at $x = 0$ (value $1$ ).

Answer: as listed.

Solved Example 27

f (x) = (x + λ)^{2}

for

x < 0

and

f (x) = cos x

for

x \geq 0

, find the possible values of

λ

such that

f

has a local maximum at

x = 0

Solution:

$f (0) = 1$ , and $cos x < 1$ just to the right of $0$ .
Just to the left we need $(x + λ)^{2} \leq 1$ . If $∣ λ ∣ > 1$ the values are near $λ^{2} > 1$ : fails. If $∣ λ ∣ < 1$ they are near $λ^{2} < 1$ : works.
$λ = 1$ : $(x + 1)^{2} < 1$ for $- 2 < x < 0$ : works. $λ = - 1$ : $(x - 1)^{2} > 1$ for $x < 0$ : fails.

Answer: $λ \in (- 1, 1]$ .

Solved Example 28

Find two positive numbers

x

and

y

whose sum is

35

and for which

x^{2} y^{5}

is maximum.

Solution:

$f (x) = x^{2} (35 - x)^{5}$ , $x \in (0, 35)$ .
$f^{'} (x) = 2 x (35 - x)^{5} - 5 x^{2} (35 - x)^{4} = x (35 - x)^{4} (70 - 7 x) = 0$ gives $x = 10$ , where $f^{'}$ changes from $+$ to $-$ .

Answer: $x = 10$ , $y = 25$ (check with the rule $\frac{x}{2} = \frac{y}{5}$ ).

Solved Example 29

A normal is drawn to the ellipse

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

. Find the maximum distance of this normal from the centre.

Solution:

At $(5 cos t, 4 sin t)$ the normal is $5 x sin t - 4 y cos t = 9 sin t cos t$ .
Distance from the origin: $d^{2} = \frac{81 sin ^{2} t cos ^{2} t}{25 sin ^{2} t + 16 cos ^{2} t}$ . With $u = tan^{2} t$ : $d^{2} = \frac{81 u}{( 1 + u ) ( 25 u + 16 )}$ .
Maximise $\frac{u}{25 u ^{2} + 41 u + 16}$ : the derivative vanishes when $16 - 25 u^{2} = 0$ , $u = \frac{4}{5}$ , giving $\frac{1}{81}$ .

Answer: maximum distance $1$ ( $= a - b$ for this ellipse).

Solved Example 30

A line is drawn through

P (1, 2)

to cut the positive coordinate axes at

A

and

B

. Find the minimum area of

△ O A B

Solution:

Let $A = (a, 0)$ , $B = (0, b)$ . $P$ lies on $\frac{x}{a} + \frac{y}{b} = 1$ : $\frac{1}{a} + \frac{2}{b} = 1$ , so $b = \frac{2 a}{a - 1}$ , $a > 1$ .
Area $= \frac{ab}{2} = \frac{a ^{2}}{a - 1}$ ; derivative $\frac{a ( a - 2 )}{( a - 1 ) ^{2}} = 0$ gives $a = 2$ (minimum), $b = 4$ .

Answer: minimum area $4$ square units (intercepts $2$ and $4$ ; $P$ is the midpoint of $A B$ ).

Solved Example 31

Find the local maximum and local minimum values of

f (x) = x^{4} - 62 x^{2} + 120 x + 9

Solution:

$f^{'} (x) = 4 x^{3} - 124 x + 120 = 4 (x - 1) (x - 5) (x + 6) = 0$ at $x = - 6, 1, 5$ .
$f^{''} (x) = 12 x^{2} - 124$ : $f^{''} (- 6) = 308 > 0$ , $f^{''} (1) = - 112 < 0$ , $f^{''} (5) = 176 > 0$ .
$f (- 6) = - 1647$ , $f (1) = 68$ , $f (5) = - 316$ .

Answer: local maximum $68$ at $x = 1$ ; local minima $- 1647$ at $x = - 6$ and $- 316$ at $x = 5$ .

Solved Example 32

The maximum value of

(\frac{1}{x})^{x}

for

x > 0

is
(A)

e

(B)

e^{1/ e}

(C)

e^{- e}

(D)

\frac{1}{e}

Solution:

Answer: (B). Maximise $ln y = - x ln x$ : derivative $- (ln x + 1) = 0$ at $x = \frac{1}{e}$ , a maximum (sign $+ \to -$ ). Then $y = e^{1/ e}$ .

Solved Example 33

Two positive numbers have sum

16

. The least possible value of the sum of their cubes is
(A)

1024

(B)

512

(C)

2048

(D)

4096

Solution:

Answer: (A). $f (x) = x^{3} + (16 - x)^{3}$ , $f^{'} (x) = 3 x^{2} - 3 (16 - x)^{2} = 0$ at $x = 8$ ; $f^{''} (8) = 96 > 0$ . $f (8) = 2 \times 512 = 1024$ .

Solved Example 34

Find the points on the curve

x^{2} = 2 y

nearest to the point

(0, 5)

Solution:

For $(x, y)$ on the curve, $D = x^{2} + (y - 5)^{2} = 2 y + (y - 5)^{2}$ , $y \geq 0$ .
$\frac{d D}{d y} = 2 + 2 (y - 5) = 0$ gives $y = 4$ ; $\frac{d ^{2} D}{d y ^{2}} = 2 > 0$ : minimum. Then $x^{2} = 8$ .

Answer: $(\pm 22, 4)$ , at distance $3$ .

Practice Questions

Find the points of local maxima or minima of (i) $f (x) = (x - 1)^{3} (x + 2)^{2}$ (ii) $f (x) = x^{3} + x^{2} + x + 1$ .Answer: (i) maximum at $x = - 2$ , minimum at $x = - \frac{4}{5}$ (ii) none ( $f^{'} > 0$ always)
Let $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6$ . (i) Find the possible points of maxima/minima for $x \in R$ . (ii) Find the number of critical points for $x \in [0, 2]$ . (iii) Find the global maximum and minimum on $[0, 2]$ . (iv) Prove that on $(1, 3)$ $f$ has no global maximum.Answer: (i) $x = 1, 2$ (ii) one ( $x = 1$ ) (iii) minimum $f (0) = 6$ , maximum $f (1) = 11$ (iv) values approach $f (3) = 15$ without reaching it
Let $f (x) = \frac{x}{2} + \frac{2}{x}$ . Find the local maximum and minimum values. Explain why the local minimum value is greater than the local maximum value.Answer: maximum $- 2$ at $x = - 2$ ; minimum $2$ at $x = 2$ ; they lie on different branches separated by $x = 0$
Find the points of local maxima or minima of $f (x) = sin 2 x - x$ , $x \in R$ .Answer: maxima at $x = nπ + \frac{π}{6}$ , minima at $x = nπ - \frac{π}{6}$
Let $f (x) = sin x (1 + cos x)$ , $x \in (0, 2 π)$ . Find the number of critical points and identify the points of maxima and minima.Answer: three: $\frac{π}{3}$ maximum, $π$ neither, $\frac{5 π}{3}$ minimum
A square piece of tin of side $18 cm$ is made into an open box by cutting equal squares from the corners and folding up the flaps. What should be the side of the square cut off for maximum volume?Answer: $3 cm$
Prove that a right circular cylinder of given total surface area and maximum volume has its height equal to the diameter of its base.Answer: $V = \frac{r ( S - 2 π r ^{2} )}{2}$ is largest at $S = 6 π r^{2}$ , giving $h = 2 r$
Towns $A$ and $B$ are on the same side of a straight road at distances $a$ and $b$ from it; the feet of the perpendiculars are $C$ and $D$ , with $C D = c$ . A hospital $P$ on the road must make $A P + P B$ least. Find $P$ .Answer: $C P = \frac{a c}{a + b}$ (reflect $B$ in the road)

Common Mistakes to Avoid

Watch out

Treating $f^{'} (c) = 0$ as proof of an extremum: $x^{3}$ has $f^{'} (0) = 0$ and no extremum. Check the sign change.
Missing critical points where $f^{'}$ does not exist: corners of $∣ x ∣$ -type functions, cusps, joins of piecewise definitions.
Calling end points critical points. They are not, but they must still be checked for global extrema on closed intervals.
Concluding 'no extremum' when $f^{''} (c) = 0$ : the test is silent ( $x^{4}$ has a minimum at $0$ ). Use the sign of $f^{'}$ or higher derivatives.
Accepting a value that is only approached (an open end or a hollow point) as a global maximum or minimum.
Assuming a local maximum value must exceed every local minimum value: $\frac{x}{2} + \frac{2}{x}$ has maximum $- 2$ and minimum $2$ .
In applied problems, optimising a two-variable expression without first using the constraint, or ignoring the domain of the variable (for the semicircle rectangle, $x \in (0, 2 r)$ ).
Forgetting to check that the critical point gives a maximum (not a minimum) in optimisation, or not comparing with the end values.

Frequently Asked Questions

What is the difference between local and global maximum?

A local maximum is the largest value of $f$ in a small neighbourhood of a point; a global maximum is the largest value on the whole domain or interval and must actually be attained. A function can have many local maxima but at most one global maximum value.

What are critical points of a function?

Critical points are interior points of the domain where $f^{'} (x) = 0$ or where $f^{'} (x)$ does not exist. Every local extremum inside an interval occurs at a critical point, so they are the only candidates to test, along with the end points for global questions.

What is the first derivative test for maxima and minima?

At a critical point $c$ , look at the sign of $f^{'}$ just before and just after $c$ . A change from positive to negative gives a local maximum, from negative to positive a local minimum, and no change means there is no extremum at $c$ .

When does the second derivative test fail?

It fails when $f^{''} (c) = 0$ (and of course where $f^{''}$ does not exist). Then use the first derivative test or the first non-zero higher derivative: an even-order derivative gives an extremum, an odd-order one gives none. For example $x^{4}$ has a minimum at $0$ although $f^{''} (0) = 0$ .

How do you find the absolute maximum and minimum on a closed interval?

Find the critical points inside $[a, b]$ , evaluate $f$ at these points and at $a$ and $b$ , and compare the numbers. The largest is the absolute maximum and the smallest the absolute minimum; no classification of the critical points is needed.

Can a function have a maximum at a point where it is not differentiable?

Yes. Extrema only compare values: $- ∣ x ∣$ has a maximum at $0$ where it has a corner, and a function can even have a maximum at a point of discontinuity. Such points are critical points because the derivative does not exist there.

How are maxima and minima asked in JEE Main?

JEE Main asks local extrema of polynomial and trigonometric functions, greatest and least values on an interval, parameters for which a function has extrema at given points, and applied problems on areas, volumes and distances. The first and second derivative tests solve most of them.

What kind of maxima and minima problems appear in JEE Advanced?

JEE Advanced asks extrema of piecewise and non-differentiable functions, global extrema on open intervals, conditions on parameters through the roots of the derivative, extrema of composite or reciprocal functions, and geometric optimisation with conics, reflections and three-dimensional solids.

Previous year questions on Maxima & Minima

32 questions from past papers, each with a step-by-step solution.

Show all 32 questions

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1. Local and Global Extrema

2. Extrema of Differentiable Functions

2.1 A necessary condition

2.2 A sufficient condition: the sign change

2.3 First derivative test

3. Critical Points and Continuous Functions

4. Global Extrema of Continuous Functions

4.1 On a closed interval [a,b]

4.2 On an open interval (a,b)

5. Second and Higher Derivative Tests

5.1 Second derivative test

5.2 The nth derivative test

5.3 Extrema of related functions

6. Applications: Optimisation

6.1 Shortest paths: the reflection trick

7. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between local and global maximum?

What are critical points of a function?

What is the first derivative test for maxima and minima?

When does the second derivative test fail?

How do you find the absolute maximum and minimum on a closed interval?

Can a function have a maximum at a point where it is not differentiable?

How are maxima and minima asked in JEE Main?

What kind of maxima and minima problems appear in JEE Advanced?

Previous year questions on Maxima & Minima

Ready to master Application Of Derivatives?

4.1 On a closed interval $[a, b]$

4.2 On an open interval $(a, b)$

5.2 The $n$ th derivative test