C ha pt er - 3

CO NC AV IT Y A N D
IN FL EC TI O N PO IN TS
3.1 : Introduction •

In our earlier class es, we have used the sign of the seco
•
nd deriv ative at a criti cal poin t to disc over
er the func tion /has a max imum or mini mum at that point. For draw ing the nece ssary conc lusio n,
wheth
appl ied the first deriv ative test. We shal l now
we rega rded /" as the deriv ative of the func tion /' and then
conclusion, which would give us an idea abou t
use the monotonicity cons idera tions off ' to draw another
the shape of the graph of the function f
mine the type ·of grap h, the given func tion
In this chapter, we shal l use the seco nd derivative to deter
rentiable in its domain.
f has. We shall assume that the func tion / is twic e diffe
whether the grap h offis risin g or falling.
We begin first by show ing how the sign of/ ' is relat ed to
tion
3.1.1 : Increasing func tion and Dec reas ing func
eas for a decr easin g function the grap h
For an increasing function, the grap h ofy = f{x) rises, wher
ofY =.f{x) falls.
Thus, we have the following definition:
Definition:
Let a func tion /be defined on an interval I and let xi' x 2
EI.
Then f is said to be
(i) increasing on I if x I < x 2 ⇒ f{x) ~f{x2) (ii) decreasing on I if x 1 < x2 ⇒ J(x) ""2::.j(x 2)
I
... ) x_
(
~n strictly increasing on I if/(x 1) <j(x2 ) whenever x 1 < 2
x < x2_
(iv) strictly decreasing on I if/(x 1) > j(x2 ) whenever 1
y y
f (xi) i---- --- f(x1) 1--............_
f (x1)i-------"
I (x2) 1 - - - - + - -___,,.

--0 ~-- ---L ---J ~-~ X

i
X1 Xi
F' 3 Fig.-3.1 (b) [f strictly decreasing]
tg.- .1 (a) [/ strictly increasing]
Example -1 : . . tl increasing, bec aus e x1' x2 E R with
x === 2x + 5, xER 1s stn c Y
• The function f defined by/ { )
X < X ⇒ 2.x + 5 < 2.x2 + S
i 2 ⇒
u;) <f{ x)
2 . t . tly decreasing on R, bec aus e xl' x2e
The function f defined by /{x) _ 2x + 5 1s s nc R with
J\: 1 ===
•
X 1 < X 2 ⇒ - 2x + 5 > - 2x2 + 5
l

⇒ N1 ) > f{x } d
· off ' for eter nun·1·ng the gra ph of f to be rising or
Now let us see the role of the sign falling
·

Fig. 3.2(a)
In fig. 3.2 (a), wh enf is increasing on (xl' x
Fig . 3.2 (b)
) the gra ph of/ wil l be incl
2 ined upw ard and the graph
has tang ent lines with pos itive slope on
(xl' xi) i.e., tan 0 > 0 i.e., f' (a) > 0
Similarly, wh enf is decreasing [see fig. 3.2
(b)], the incl inat ion of the tan gen t toy =
poin t Pis an obtuse angle and tan 0 < 0 i.e., ./{x) at any
f' (a) < 0.
Thu s, it is reas ona ble to exp ect f to be incr
easi ng on I if J' (x) > O and dec reas ing on I if
f' (x) < 0, for allx EI.
Hence, we hav e the following theorem:
Theorem - 1 : [Monotone Function Theorem
]
Let /be differentiable on the interval I= (a,
The n b).

(i) f' (x) > 0 on I ⇒ f is strictly increasing on I.

(ii) f' (x) < 0 on I ⇒ f is strictly decreasing
Pro of: on I.
(i) Let x 1, x 2 be two distinct points in (a b) and
Sin ce/ is differentiable on (x x) th,er ti
x! < x2.
we can get a number c in (a, b) " 2 e ore f 1s co f
such that x < c < n inuo us on [xl' x2]. By
and 1 X
Me an Val ue TheoreJl1'
2
/'(c )= /(x 2)- f(x1 )
X2-X 1
⇒ f{x2) - j{x1) = f'(c ) (x -x )
2 1
... (1)
eoncavity and Inflection Points
3.3
Since/'(x) > 0 for allx E (a, b), therefore f' (c) > o
Also, x2 -x1 > 0
From ( 1), we have
.f{x2) - ./{x 1) > 0
⇒ .f{x1) <.f{x2) i.e., if xi' x2 E (a, b) such that x 1 < x , then.f{x ) </{x ).
2 1 2
Hence f is strictly increasing on (a, b)
Proof of (ii) is similar.
Note :If/' (x) = 0 in (a, b), then/is constant in (a, b).
In view of Theorem 1, we have the following conclusion:
To determine, whether a function/ is increasing or decreasing, we have to find the numbers where
f' is zero or does not exist. These numbers divides x-axis into intervals and we have to test the sign
off' in each of these intervals. If
( 1) f' (c) > 0 in an interval, then f is increasing in that interval and
(2) f' (c) < 0 in other inerval, then/ is decreasing in that interval.
Hence c is called critical number, where /' (c) = 0 or /' (c) does not exist.
Problem - 1 : Find the intervals in which the following functions are (i) increasing (ii) decreasing.
X
(a) .f{x) = x 2 + l (b) fix)= et+ e-x

, - (x2 +1)-2x2 l-x2 (l+x)(l-x)

Solution: (a)/ (x) - (x2 + 1)2 - (x2 + 1)2 (x2·+ 1)2

f' (x) = 0 ⇒ 1 -x2 = 0

⇒ x = -1, 1 are the critical numbers.

I
-------~~------?
· -1\ Jl
x-axis
I'-..___../
The critical numbers divides x - axis into three parts. i.e., (-oo, -1), (- 1, 1) and (1, 00)

Ifx <-1 i.e., x E(-oo, -1), then/' (x) < 0.

[For x = -2 f' (x) = (-2 + l)(l + 2) < 0]

' 5
If x > 1 i.e., xE (1, oo), then /' (x) < 0

[For x = 3, f' (x) = (1 + 3 ig- 3) < 0]

If-1 <x < 1 i.e., x E(-1, 1) then/' (x) > 0

[For x = 0, f' (x) = 1.1 = 1 > 0]
 C r
J3~
.4~_ __ ; _ ~ _ : _ - - - - - - - - - - - - - - - : : : ~ f
Hence
(i) /is increasing on (-1, 1)
(ii) f is decreasing on (-00, - l)u(l, oo) f'>O
Remark:
Drawing the graph off and/', we can see that f' <O
( i) The critical numbers off are x = - l and x = 1 a nd they
are the x-intercepts of the graph off' · . x < - 1 and x > 1.
(ii) The graph of/is rising when -1 < x < 1 and falhng when
The graph off' is below x-axis. -1 1
(b) ./{x) = e + e-x

f' (x) = 0 ⇒fi1-x = 1 ⇒ X = 0

x < 0 ⇒ f' (x) < 0 andx > 0 ⇒ f' (x) > 0
f is increasing on (0, oo ) and decreasing on (--oo, 0)
Now we study the behaviour off' near a criticial number, which is called the first derivative test
for relative extrema.
3.1.2: First derivative test for an extremum
Step -1 :
Find the critical numbers c for the continuous function f, such that f (c) is defined and either
f' (c) = 0 or f '(c) does not exist.
Step - 2:
(i) The point (c,./{c)) is a relative maximum if/ '(x) > 0 for all xE(c - h, c), h > O
andf'(x) < 0 for all xE(c, c + h), h > 0.
(ii) The point (c,./{c)) is a relative minimum if f'(x) < O for all xE (c _ h, c), h > O
andf'(x) > 0 for all xE(c, c + h), h > 0.
(iii) The point (c,./{c)) is not an extremum if f'(x) has same sign for all x E(c _ h c)
and (c, c + h) on each side of c. '
Problem-2:
Using first derivative test, find the relative maximum and 1 -
re a11ve nnrumum 1 any, at e given
.. . • . 'f th
. cntlcal numbers, of the function/ defmed by /{x) = (x2 _ 4)4 (x2- l)3 at x = x = _
Solution : We have · 1' 2
j{x) = (x2 - 4)4 (x2 - 1)3 = (x + 2)4 (x - 2)4 (x + 1)3 (x _ 1) 3
3
f'(x) = 2x (x + 2) (x - 2)3 (x + 1)2 (x - 1)2 [3 (x + 2) (x - 2) + (
= 2x (x2- 4)3 (x2 - 1)2 (7x2 - 16) 4 x + 1) (x - l)]

In this case, if xE(l- h, 1) i.e., for all x in the interval to the ft

1e of 1 or XE(l, 1+h) 1.e., for a x in
. 11
the interval to the right of 1.
case -I:
f'(I - h) > 0 and/ '(l + h) > O
side of I.
:. f' has same sign for all x in (I - h, I) and (I, I + h) on each
:. (IJ{ l)) i.e., (I, 0) is not an extremum.
case -II:
/'(2 -h) < 0 and/ '(2 + h) > O
:. The point (2,/ (2)) i.e., (2, 0) is a relative minimum.
3.2 : Concavity and inflection points
derivative changes sign.
We have seen that a local extremum is a point where the first
derivative changes sign;
. In this section, we shall discuss points, where the second
al. In other words, as x increases,
If/" > 0 in some interval [a, b], then /' increases in this interv
the tangent turns counter clockwise. This
the slope of the tangent to the graph at (x,J{x)) increases, so that
results in the graph bending upward or bending downward.
s on graph which corresponds to
Such graphs lie below their chord (line segment joining the point
x = a and x = b) and above their tangents. The graph of/,
in this case, is concave upward on a< x < b, if
/'>0 on this interval. [See fig. 3.3(a)].
slope decreases, the tangents turns
Similarly if/" < 0 in [a, b], so that f 'decreases and hence
and above the chord. [see fig. 3.3 (b)]
clockwise. The graph off, in this case lies below the tangents

y
y

-4
0
-- -- -- -- ~x
. - 3•3 (a) Fig. - 3.3 (b)
Fig.
Definition - 1:
If/ is differentiable on the interval (a, b), then the graph of/ is
al
(a) concave upward on (a, b) if/'i s increasing on the interv
(b) concave downward on (a, b) if/'i s decreasing on (a, b).
Definition - 2:
graph of the function f is above the
A function f is said to be concave up on an interval if the
. .
tangent at each point of the interval. graph of the funct10n below the
/ 1s
A function/ is said to be concave down on an interval if the
tangent at each point ~f the interval.
(b)]
Consider two increasing functions [see fig. 3.4 (a) and fig. 3.4
 y y

- .i-- - - - --~ x
0
Fig.-3.4 (a) Fig.-~.4 (b)
In case of the two functions above, their concavity relates to the r~te of ~~rease .
Since both are increasing, the first derivative of both functions ts pos.ittve. The first function
[fig. 3.4(a)] increases at an increasing rate, because the slope of the tangent line becomes steeper and
steeper as the values of x increases.
So the first derivative of the first function is increasing. Thus, the derivativ e of the first derivative
'
i.e., the second derivative is positive. This function is concave up. '
The second function [fig. 3.4(b)] increases at an decreasing rate, so that the first derivative of the
second function is decreasing, because the slope of the tangent line becomes less and less steeper as x
increases. So the derivative of the first derivative i.e., second derivative is negative. This function is
concave down.
Similarly, let us consider two decr~asing functions [see fig. 3.5(a) and fig. 3.5 (b)]

y y

Fig.~3.~ (a) . Fig.-3.5 (b)

The first decreasmg function [3.5(a)] decreases at an increasing rate. In this case the slopes of
tangent lines becomes less and less negative as x increases.
So, the first derivative of the first function is increasing. Thus the deri t·
. . . .. h'
i.e., the second denvatI~e 1s p~s1tive. T .ts function is concave up. , va IVe o fth e firrs t derivative
T~e second function [Ftg. 3.5(b)] ?ecreas~s at an decreasing rate. In this case, the slope of the
tangent lme becomes more and more negative as x increases So the firrst d · t·
. . . · ,
is decreasing. Thus, the denvative of the first denvative i e second deri· enva 1ve o f t he second function
· ·, vat·1ve 1s
. negative . . This function
is concave down.
Hence, the concavity corresponds to the rate of change of the first d • t· . .
. . . . .
the sign of the second denvative by mcreasmg/decreasmg test applied t enva 1ve 1.e., 1t corresponds to
th d . .
o e envative. ·
concavity-itnc{ lnftedion 'Points
s of concavity
_2.1: second derivative test for determining interval
3
Step-I: a
the values of x for whi ch[" (x) = 0 or f" (x) does not exist, and mark these numbers on
Find
ber of open intervals.
number line. These numbers divides the line into a num
Step- 2:
step 1 and evalaute f" .
Choose a number c from each interval determining in
Then,
a< x < b.
(i) if f" (c) > 0, the graph off is concave upward on
on a< x < b.
(ii) if f" (c) < 0, the graph of/i s concave downward
. . 1 f · for the function f defmed by j{x) = -l x 3 - 9x + 2
avity
Problem - 3 : Detenmne mterva s o conc 3
Solution : We have
1
2
./{x) =
3x3- 9x+
f' (x) =x2 - 9 and f" (x) = 2x
f" (x) = 0 ⇒x = 0
Take the points -1 and 2.
We find f" (-1) =-2 < 0 and f" (2) = 4 > 0
Thus f" (x) > 0 if x > 0 and f" (x) < 0 if x < 0
ave down on (- oo, 0).
So, the graph off is concave up on (0, + oo) and conc
: In the interval (a, oo),
Thus, for concavity test, look at the following example
Concave down Concave up

a
L.J.~ b
Fig. - 3.6
and concave down on (b, oo).
the function shown in Fig. 3.6 is concave up on (a, b) b))
The point at which a function is changing concavity is called the inflection point. Hencce (b,./{
is the inflection point.
Definition - 3 :
of/, where the concavity changes i.e., the
~ inflection point is a point P(c, /{c)) on the graph
n on another side of P.
graph Is concave up on one side of p and concave dow
.
At such a point, either f" (c) = O or f" (c) does not exist
point c is such that, f" (c) is either zero or undefmed , then c is the critical point off ' . We call
If a
that f' and f" exist, then we have the following
such c as a second order critical number. So iff is such
three cases:
Case-I·
• . • · e This function is concave down befi
The sign off" is changing from negative to pos ittv · ore c
· t t x = c [see fig. 3.7 (a)].
1
. .
concave up after c and has an mflect10n pom a
-1-
f"

I C
Concave up
Concave down
Inflection at c
Fig. - 3. 7(a)
Case-II:
The sign off" is changing from positive to negative. This function is concave up before c, concave
down after c and has an inflection point at x = c. [see Fig. 3.7 (b)]
f" +

/ Concave up c Concave down

Inflection at c
Fig. - 3. 7(b)
Case-ill:
If the sign off" is not changing at x = c, then/ does not have an inflection point at x = c.
[see Fig. 3.7 (c) & Fig. 3.7 (d)]
+ +
f" f"

I Concave up
C
Concave up f Concave down c Concave down
no inflection at c no inflection at c
Fig. - 3.7(c) Fig. - 3. 7(d)
Example- 2:
• Let.f{x) = x4
f'(x) = 4x3 and f" (x) = 12x2
Now f"(x)=0 ⇒ x=0

So, 0 is a critical point off'.

f" (x) > O for all x '# 0, hence the sign off" does not change in passing through o.
Hence, there is no inflection point at x = O.
Looking at the graph ofj(x) = x4, we see that it is concave up in (-<X) ,<X> ). [See Fig. 3.8]
~
, concavity and Inflection Point s 3.9
y

4
0 f (x) = x :/has inflection at (0, 0)

Fig. - 3.8
of J' .
Note : A function does not necess arily have an inflection point at a critical point
Exam ple-3:
I
• Letj{x) = - ,x E R - {O}
X

I 2
f'(x) = -z, f" (x) =-3
X
X
Clearly
(i) f" (x) > 0 if X > 0
(ii) f" (x) * 0 for any xER- {O} and
(iii) f" (x) < 0 if X < 0
exists for all
(i) implies the graph of/ is concav e upwar d in (0, oo ). From (ii) and the fact that/"
x in the domai n of/, we conclu de that the graph has no points of inflecti
on. (iii) implies that
the graph of/ is concav e downw ard in (-oo, 0). [see fig. 3.9]
y

x' ~- --- --+0--- --- ~ x

Y'
Fig. - 3.9
- CQl rJ
3.10 ~I
3.2.2 : Procedure for finding inflection points
Step - 1:
Compute/' (x) and find all points in the domat·n off, w here e1t· her /" (c) = 0 or
/' (c) ~ n~
exist.
Step- 2: . .
For each number c obtained in step I, deternune the sign off " to the left of x = c and to the .
of x = c. If/' (x) > 0 on one side and/ ' (x) < 0 on the other side, then the point
(c, J{c)) ::&In

Remark:
inflection point off
~
If f' does not change sign at c, then f does not have an inflection point at (
c, j(c)).
Problem - 4 : Determine the concavity and the inflection points of the functi
on f defined by
j{x) = 3x4- 4x2 + 1
Solution : ./{x) = 3x4 - 4x2 + 1
:. f'(x) = 12.x3 - 8x and f" (x) = 36x2 - 8
To find inflection points, we have to find second order critical numbers wher
e/" (x) = 0
Now, f'(x)= O ⇒ 9.x'-2=0 Is
2 12.
⇒x2 =- ⇒x =+-"-L
9 - 3
Concave
Concave up I down Concave up
E I I · I I 1. I )
2 -l -0.47 0 0.47 1 2
Fig. - 3.10

Let us select a convenient number less than - .fi, ~

Jiand greater
3 (~ -0.47 ), between - - and- ✓2
~ 3 3
than3- .

Take the numbers -1 ' 0, 1 respectively

.
/"(-1 )=36 -8>0 .
/"(0) =-8 < 0
f" ( 1) = 36 - 8 > O

f" (x) > 0 if x < _ ✓

3
2 and
x>✓2-
3
and /" (x) < Oif - ✓32 < <✓
X -
2
3
concavi.ty an d Inflection Points 3.11

The gra ph of/ is con cav e up for values

of x less than - ../2 and greater than✓2- and concave down
3 3
.fi .fi
between -
3 and
3 .
t. l . fl . · t at - - and - ../2 ../2
e the concavt·ty changes.
Thus, th e fun c ion 1as m · ectton pom , si·nc
3 3
inal fun ctio n/{ x), we have
Putting the se two val ues into the orig

+f]=3(- 4
f J- (- f J +I =~ ~- : +I=::= ;1
an d{ f]=3( f J - 4( fJ +I = ;
So, ( - f' ;7] f' 2~]
and ( are inflection points.
of the cur ve y = (x + 1) tan-Ix
Prohlem - 5 : Fin d the poi nts of inflection
Solution ; We hav e
y = (x + 1) tan - x =j{ x) (say )
1

1 x+ l
+ tan-Ix = + tan-Ix
f' (x) = (x + 1). 2
l+x
2
l+ x
(1- x ~
2 2
"_ ( 1 + x ) - ( x + 1) 2x 1 _
f (x) - 2 + 2 - (x 2 + 1)
(x2 + 1 ) l+x

To find critical poi nts off ' .

So put f" (x) = 0 ⇒ x = 1
ber line. Take -1 and
1 on the num ber line and to test bot h intervals to which it divides the num
Put
2 as the test points.
I
ll
pos itiv e neg ativ e

1
Fig. - 3.11
We fmd
2
f'(- 1) = (1 + I) =I > 0 and /" (2) = ;~ < 0 ,
4
f" (x) > 0 if x < 1 and /" (x) < 0 if x >
1
00 )
d/ is concave down on (1,
He nce / is con cav e up on (-oo, 1) an

(I ,f(I)) is the inflection point i.e., ( I, ;

) is the inflection point.
 -¾&7: $.,

':eilcti:g
3.12 ~
h O f.ft )- ax3 + bx2 + c has an inflection po·
Problem - 6 : Determine a, b and c such that the grap x - lllt and
slope 1 at (-1, 2)
Solution : We have
j{x) = ax3 + b.x2 + c
:. f' (x) = 3a.x2 + 2bx and /" (x) = 6ax + 2b
The graph of/has slope 1 at (-1, 2)
i.e. , f' (-1) = 1
... (1)
⇒ 3a-2b= 1
The graph of/has an inflection point at (-1, 2),
So/" (-1) = 0
⇒ - 6a+2b=0
... (2)
⇒ -3a+b=0
1
Solving (1) and (2), we get, a = -
3, b = - 1
Since an inflection point must be on the graph, therefore
/{- 1) must be defined and
j{-1)=2 ⇒ -a+b+c=2 ...(3)
8
Substituting the values of a and b in equation (3 ), we get c =
3
1 8
Hence a= -- b = - 1 c = -
' 3' ' 3
Problem - 7 : Find the points of inflection for the function f defined by
j{x) = x 4 + 4.x3 - 18.x2 + 9x- 3.
Solution: We have/'(x) = 4.x3 + 12.x2- 36x + 9 (1

f "(x) =12x2 + 24x -36

At the points of inflection, f "(x) =0
⇒
2
12x + 24x-36 =0
⇒ x=-3, 1
If x=-3, f(x)=-319
If X = 1, f (X) = - 7
:. Points of inflection are (- 3, - 319) and (1, _ 7).
Here/" (x) changes sign from negative to positive sigll
from positive to negative as x crosses through_ ~s x crosses through 1 and/" (x) changes
3
Problem - 8 : Examine each of the following functions fi . . foO·
1
a) f(x) = x3 or concavity, convexity and points of inflec
( (b) f (x) == x113
(~) f(x) = X 4 -2x3 -12x2 + I (d) f(x)==:~
(e) f(x)=fnx, x>O ' x-3
concavity and,n]lection'points ·
-
Solution :
(a) fi.x) = x3
f' (x) = 3x2
f' (x) = 0 ⇒ x = 0 and J" (x) = 6x
f" (0) = 0 .
fm(x) =6,fm(o) = 6 -:t= 0
.·. x = 0 is a point of inflection.
If x > O,f" (x) > 0
⇒f is concave upward
:. /is concave upward in (0, oo)
If x < 0 i.e., x E (- oo , 0), then/"(x) < o
:. /is concave downward in (- oo, 0).
I
(b) f(x) =x3
2
.
.. f' ( X ) -- .!.. X -3 -
- 1- -
1
213
3 3· x
.·. x = 0 is a point of inflection
"( X )- 1 -2 -¾
--X-X
_
----
2 1.
/ 3 3 9. XS / 3
/" (x) > 0 if X E (-oo, 0)
.·. /is concave upward in (-oo, 0)
f" (x) < 0 if x E (0, oo)
.·. /is concave downward in (0, oo)
(c) f(x) = x4-2x3 - 12x2 +7
f' (x) = 4x3 - 6x2 - 24x
f"(x)= 12x2 - 12x-24= 12(x2 -x-2)= 12(x2 -2x+x-2)= 12(x-2)(x+ 1)
f"(x)=O ⇒ x=-1,2

f"'(x) = 24x-12
/"'(-1) -:t= 0 &f"'(2) -:t= 0
.·. -1 and 2 are points of inflection.
f" (x) > 0 if x E (-oo, -1) u (2, oo)
.·. f is concave upward in (-00 , -1) u (2, oo)
/" (x) < 0 if -1 < X < 2
.·. f is concave downward in (-1, 2)

(d) / ( X) = X- 2
x-3
, (x-3)-(x-2) _ -1
f(x)= (x-3}2 -(x-3}2
_13~.1!_4_ _ _ _ __:__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _~1~ ~

f '(x) * O for any vlaue of x. There IS• no Point of inflection. "-

2
f" (x) = (-1) (-2) (x-3)-3 = 3
(x-3)
f" (x) > 0 if X > 3
⇒ f is concave upward in (3, oo)
f" (x) < 0 if X < 3
⇒ f is concave downward in (-oo, 3)
(e) .f{x) = In x, x > 0

:. f'(x)=..!.... f' (x)-:t:-Oforanyrealx.

X
.·. There is no point of inflection.
1
f"(x) = - - & f" (x) < 0
x2
⇒ f is concave everywhere i.e., concave downward.
Problem - 9 : Find the points of inflection on the curve y = ax3 + bx2 + ex + d. Also state the conditi~
that the co-efficients a, b, c satisfy for the given curve to have points of inflection.
dy d2
Solution : We have - = 3ax 2 + 2bx + c and - { = 6ax + 2b = 2(3ax + b) .
dx dx

⇒ 3ax + b = O
2
The given curve has points of inflection if and only if d -; = 0
dx
d 2y -b d 3y
--
2 =O ⇒ x=-,a-:t:-Oand dx 3 =6a
dx 3a
d3
-{=6a -:t:-0
dx
The curve has points of inflection if a * 0 E R and b, c can have any real value.
If x=-~, then y =a(-~)+a(-~)+a(-!?_)+d
3a 27a 9a 2 3a
b3 b2 b 2b3 b
=----+---+d
3 2 =---+d
27 a 9a 3a 27 a 2 3a
3
b 2b b )
: . Point of inflection is ( - a, a2 - a + d
3 27 3
Problem - 10 : Find the points of inflection of the curve x =at t _ .
an , Y - a smt cost.
dx 2
Solution : We have dt = a sec t

dy [. ( . )
dt = a smt -smt +cost.cost)]== a(cos 2 t- sin2 t)
concavity an d lnfleef[on Points
2
~~.:___ _ _ _ _ _ __ ~ - -- -- _ :__;3!,.,!_15~
__::.---~-~d;~~~dy;/~dx;_~c~o:s2~t~s:in 1 == 4 .
dx dt dt cos t - sm2t . cos2t
2
sec t
-

2
- cos t (cos2t - sin2t)
-

- cos 2 t(2cos 2t _ 1)
-

-- 2 co st4
- cos2t
d 2y d
d (2 4 d
2 = dx (2cos4 t-c os 2t) == dt cos t - cos 2t)_,_ t
dx
dx
== -8 co s3tsi nt+ 2c os tsi nt
asec 2t
1
== a [2cos3 tsi nt- 8c os s tsint]

1 3
== a 2cos tsi nt( l-4 co s t)
2

2
For points of inflection' -d=y
O
dx 2
2t =I
⇒ sin t = 0 or cost = 0 or 4 cos

:. t = 0 and co st = ± ..!_
2

,x=atant=a ✓sec t-I =± ✓3a

2
Fo rt= ±~

_. RI
Y- as mt .co st= ±a l- -.
4 2
-= ± --
4 .
✓3a

Fo rt = 0, we have x = 0, y = 0

. .
The pomts of inflection are (0,
( r:;
0) and ±v3a,± 4
.fia)
d 2y . points.
that dx changes sign as x passes through each of the
It can be seen 2

3·2·3 : The se co nd derivative test

not, at a
be used also to det erm ine if there is an extreme value or

:Ic~I
. . Concacity of the functi on can Similarly a
wi th relati ve mi nim um is con cave up on an interval around it.
po~t of f A function und it.
h a relative ma xim um is concave down on an interval aro n is concave
ction wit
al po int and the sec on d der iva tive at c is positive, i.e., the functio
Thus, if c is a critic the second
is a relative mi nim um at c. Co nversely, if c is a critical point and
up _around c, then there
n is con cav e dow n aroun d c, then there is a relative maximum
functio
derivative at c is negative i.e., the
at c If f" ( )· 0, the test 1s . onc1us1.ve.
. mc
· c =
 ~!_::_~
~-16 :,, --~
2 - _ ~ _ : _ . : _ _ - - - - - - -.-. - - -i./. :~-:..-..:.-~', Cqz I~~~,Al~ti:
. open interval contatrung x = c and/' ( .d/
Let/be a function such that/" (x) exiSt S on an _ c) :::: o. ~Je'"
If/" (c) > O, then/has a relative minimum at x = c. t , j'(
1

If/" (c) < O, then/has a relative maximum at x = c. . . ~/f

1 JP
If/"(c) = 0 or f"(c) does not exist, the test fails i.e., mconclus1ve and/may have a re!a ·
· · · .• 1 t've extremum at all at x = c. tive
maxl.lllum a relative nunnnum or no re a i
' · Concave down
maximum at c
~o?cave up
V c

j,f
ar
a b C
1,1
Fig.-3.12
Problem - 11; Determine the extreme values of the following functions using the second derivative test :
s
(a) f (x) = (x2 - 3x + l)e-x F
1 1t 1t 2
(b) f(x )=sinx+ cos2.x,at x= ,x=
2 6 2
Solution : (a) f(x) = (x2- 3x + l)e-x
2
:. f' (x )=-(x2-3x + l)e-x + (2.x-3)e-x = e-x (2.x- 3 -x2 + 3x- l)= e-x (5x-4-x ) =- e-x (.x2-5x+4)
f" (x) = e-x (x 2 - 5x + 4)-(2.x- 5) e-x= e-x (x2 - 5x + 4 - 2x + 5) = e-x (x - 7x + 9)
2 )
Prob~m- '
Solving /' (x) = 0, we find critical numbers.
:. f' (x) = 0 Solution:
⇒ x 2 - 5x + 4 = 0 ( ·: e-x * 0)
⇒ x= 1, 4

To apply second derivative test, we evaluate/ " (x) at each critical number
f" (I) = 3£: 1 > Q
f " (4) = -3e-4 < 0
.·. /has a relati_ve minimum at x = 1 and f has a relative maximum at x = 4
Hence the maiimum value off is/(4) = 5e--4 and minimum value of/ is/(1) = - e-
1

(b) f{x) = sinx + cos 2x

2
: . f' (x) = cos x - sin 2x and /" (x) = _ sin x _ 2 cos 2x
Now,

/" (n)2 = - 1 + 2 = 1 > 0 and /" ( n) = _ _!_2-1--2

6
_ 3
<0

:. /has a relative minimum at x = n and 1 t· . n

.. 2 re a ive maximum at x = 6 .

. ly
The maximum and minimum value of/are respective
.
i(n)
-6 and i(nJ. e , -43 ando.
-2 i..
 , • I'-

· .. 3.'f f
. . - -· ·
bleJJl-12: Fmd_constants a, ban d c suc h that the
of.l{x) ==x + ax +bx + c will have a relative
3 2

pro maximum at (-3 , 18) and a relativ ~~p h

. f' (x) = 3x2 + 2ax + b e tnmimum at (1, -14 ). Also find.l{x).
.
sotunon.
/ha s extr emu m at x = - 3, 1
:. f' (-3 ) = 0 and /' (1) = o
⇒ 27 - 6a + b = 0 ... (1)
and 3 + 2a + b = 0 ... (2)
Also.l{-3) = 18 and /{l) = - 14
⇒ -27+9a-3b+c= 18
i.e., 9a - 3b + c = 45 ... (3)
and 1 + a + b + c = - 14
i.e., a+ b + c = - 15 ... (4).
Solving (1) and (2), we get a= 3, b = - 9
From (3), we hav e
27+ 27 + c=4 5
⇒ c=-9
The values of a, b, c satisfies (4)
:. a = 3, b = - 9 and c = - 9
.f{x) = x 3 + 3x2 - 9x - 9 e relative
h that the function/{x) = Ax + Bx + C will hav
3 2
Problem - 13: Fin d con stan ts A, B and C, suc
at (1, 5).
extremum at (2, 11) and an inflection point
Solution : f' (x) = 3Ax2 + 2Bx
f"(x ) = 6Ax + 2B
[ha s extremum at (2, 11)
⇒ ['(2 )=0
⇒ 12A + 4B = O ⇒ 3A + B = 0 ⇒ B = - 3A
.. · (1)
[ha s an inflection poin t at ( 1, 5)
⇒ f"( l) =0
⇒ 6A+2B=O .. .(2)
⇒ 3A+B=O ·
• fy · n.
From (1) and (2) B = - 3A curv e their coo rdin ates must satls its equatio
· ' h · t the ,
Smce (2, 11) and (1, 5) are t e pom son
Thu s,/ (2) = 8A + 4B + C
= 8A -12 A+ C ... (3)
=-4 A+ C= 11
and f(l) =A + B + C
=A -3A +C ... (4)
=-2 A+ C= 5
3.18

Solving (3) and (4) we get,

A=-3, C=-1 .
:. B=9
:. .f(x) = -3.x3 + 9x2 - 1
Problem -14: Find the critical numbers for the function.f(x) = tan-' (: )-tan·' (:) a> b.
Classify each as corresponding to fia relatived maximum,
. . 1 a relative minimumorn .
~,~
.
o ution: By critical number, we always mean a rrst - or er cnttca number.
S I

Given/(x) = tan·'(: )- tan- 1

(:), a> b.
b
: . f, (x) = 1 •_!_ _ 1 2 1 _ 2a 2 x2 +b2
x2 a x ·b - x + a
1+-
a2
1+-
b2
f' (x) ==0
⇒ ax2 + ab 2 - bx2 - ba2 == O
⇒ x2 (a - b) = ab (a - b)
⇒ x2=ab [·:a-b=t=O]

⇒ x =±..Jab
f" (x) = -2xa + 2xb
. (x2 +a2)' (x2 +b2)'
f" (..Jab) (-2a..Jab 2
+ 2-.Jab 2b
2
ab+ a ) (ab+ b )2 Problen

= -2a..Jab + 2b..Jab -2l + 2~

a2 (a+ b )2 b2 (a+ b )2 - y b >0 since a > b
(a+b)2 '
Solutio
Hence/has a relative minimum at x = .j;;b·
/" (-..Jab) < 0
⇒ f~s a relative maximum at x = -.j;;b.
Problem - 15 : Fmd the extreme va1ues of.f{x) •
Solution : .f{O) = 2 implies that the fu . , given that .f{O) = 2, /' (0) = 0 /" (x) > 0 for all t•
, . . ncttonf passes thr . '
f (0) = 0 implies that O is a . . ough the pomt (0, 2).
,, . . cntical number and
f (x) > 0, for all x, implies that f' tangent at (0, 2) is horizontal.
So, f" (0) > o. is always concave up.
1 eon.cavity dnd Injledtion POints
.
3.19

Hence by sec onI d de riv ati ve tes t ' th e fu nc tto n/h as a mu..umum at O.
.
. . llumva ue 1s .f{O) = 2.
:. m.t1lll
ic function Y = 2 *
rr ob lem -16 : Fo r the qu ad rat ax + bx + c (a O) ' a, b and c are constants, show that the
b
ver tex (re lat ive ex tre mu m) oc cur s at x =- _2 .
a
Q~ . Solution : We ha ve
\ y= ax 2+ bx +c

y' = 2ax + b = 0 if x =_ ~
2a ·
As sum e a> 0
b
y' < 0 for x < - -2a ⇒ y de
cre ase s

b
y' > 0 for x > - - ⇒ y inc rea ses
2a
b
oc cu rs
:. At x = - -2a , mi nim um va lue
Fu rth er ass um e a < ()
b
y' > 0 for x < - - ⇒ y inc rea ses
2a
b
y' < 0 for x > - - ⇒ y de cre ase s
2a
urs at x = - _.!:._
, xim um va lue oc cu rs. Th us, relative ext rem um occ 2a
:. At x = - 2 b ma
a
d
b) = nR T, wh ere P, V, T are the pressure, volume an
(P +; 2) (V -
Problem - 17 : Fo r the eq ua tio n nts, find T = Tc• the
of a ga s res pe cti ve ly an d a, b, n, R are positive consta
tem pe rat ure ssu rePc=P(Vc) in terms of a, b, n
po int wh ere P" (V ) =0 an d als o find the critical pre
an dR .
a
. : P(V ) = nR Tc _ _
Solution 2
V -b V
nRTc 2a
P'( V) = (V-b)z - yi

nRTc 2a ... (1)

P'( V) = 0 ⇒ (V-b)2 = y3
JI 6a
or p 2nRTc
P" (V ) = (V -b / y4

nRTc 3a ... (2)

P" (V ) = 0 ⇒ (v -b / =y4
 From (1) and (2), ·we get

·v-b= -V ==> 3V - 2V== 3b ⇒ V=== 3b

"'-,; . 2
3 . h lume that satisfies P'(Vc) = 0 and P"(Vc) == 0
The critical volume V c is t e vo
:. Ve= 3b . .-
nRT a nRTc · a
:. P(Vc)= 3b-~ - 9b2 = 2b 9b2
nRTc . _ 3a
. P" (Vc) = 0 ⇒ (V - b )3 - y4
nRTc 3a
⇒ SlT= 8lb4
⇒ 8a = 27 nb RTc
8a
⇒ Tc= 27nbR
Problem - 18 : Fm: the function/,

. -·x2 - 7x + 6
given / '(i}= - - - , f (x)
,, x2 +6x- 27 .
( )2 andj(- 3) 1s not defined.
" x+3 x+ 3
Determine the intervals on which / increases / decreases, concave up/concave down aoo
the values of x for which / has maximum, minimum and inflection.
x2 -7x+ 6 (x-l)( x-6)
Solution : f '(x) = - - - = ----
x+3 x+3
f'(x)=O ⇒ x= 1,6

Hence critical points off are 1, 6, -3.

-3 1 6
f '(x) > 0 if x > 6 and -3 < x < 1
⇒ fis increasing on (6, oo) and (-3, 1)
Fig.-3.13 (a)
f'(x)< 0ifx< -3 and 1 <x<6
⇒ /is decreasing on(~ , - 3) and (1, 6)

. at .x = 1, there is maximum and at x == 6 , there 1s

Thus · nunnnu
· · m.
Smee / 1s not defined at x = - 3 so there is no extr
, eme va1ue at x = -3.
-I +
I+
-3 1
/ l .1 l
6

decreasing
mcreasing decreasing 1
increasing
Fig.-3.13 (b)
 ctioTJ, Points
. concavity'0:nq. lnfle_
--:.: ..::- -~-- ----- ~--- =--- ----- ~~~ ----- ~3~. 2!:1,
Further,
(x+9)(x -3)
f" (x) = 0 ⇒ (x + 3)2 =0
⇒ x=-9, 3
So, the critical points off' are 3, -9, -3
f" +
-9
I I+
-3 3
f~
Concave
\u
\ fConcave
fConcave Concave
up down down up
Fig.-3.13 (c)
f" (x) > 0 if x > 3 and x < -9
⇒ /is concave up on (3, oo) and (--oo, -9) ·

f" (x) < 0 if -9 < x < -3 and -3 < x < 3

⇒ fis concave down on (-9, -3) and (-3, 3)
Thus, there are inflection points at x = 3 and x = -9 and no inflection at x = -3.
Problem - 19 : Given the properti es of a function / given below. Determine the minimum and maximum
values ofj(x), inflection points, intervals in which/ is concave up/down.
1(3) = 1,j(-3) = -1,j(0) = 0
/'(3) = 0, f'(-3) = 0, f" (x) > 0 on (- oo, 0) and f" (x) < 0 on (0, oo)
Sketch the graph of any one possible function with these properties.
Solution :/(3) = 1,j(-3) = -1,j(0) = 0
⇒/ passes through the points (3, 1), (-3, -1) and (0, 0)
Further,
/'(3) = 0 and/'(-3 ) = 0
⇒ (3, 1) and (-3, -1) are critical points.
00)
Also, f" (x) > Oon (- oo, 0) and f" (x) < 0 on (0,
This shows that we have to use second derivative test to maximum
determine if the~e points are points of minimum, maximum or neither.
f" (x) > 0 on (- oo, 0)
Since-3 e(- oo, 0), therefor e f" (-3) > 0
= -3 ( 0 , 0 )
⇒ /has minimum at x
inflection
Further, f" (x) < 0 on (0, oo)
Since 3 e(O, oo), therefor e f" (3) < 0 minimum
Fig.-3.14
⇒ /has maximu m at x =3
Hence, f" changes sign at x =0
 3.22

So, (0, ff\ is an inflection point. (0 oo)

,
f is concave up on (- oo, 0) and f ts
. ncave down on '
co 3
.h h
Problem - 20 : Find the value( s) of a for whic veftx) == x4 + ax + -x2 + 1 is
3 . concave
1
· t e cur 2 a 0ng the
number line.
3
Solution : .f{x) = .1A + ax3 +
2x2 + 1
/'(x) = 4x3 + 3ax2 + 3x
/" (x) = l 2x2 + 6ax + 3
The curve is concave if/" (x) ~ 0 for all x
⇒ 4x2 + 2ax + 1 ~ 0 for all x.
But ax2 + hx + c ~ 0 if b2 - 4ac ~ 0
Hence, 4x2 + 2ax + 1 ~ 0 if 4a2 - 4.4 ~ 0
⇒ 4a2 - 16~0
⇒ a2~4
⇒ lal~2
:. a E[-2, 2] .
Problem - 21: Find constants a, band cso that the graph offt x) = ax,2 + bx + c has a relative maximum
at (5, 12) and crosses y-c1.xis at (0, .3). Also find.f{x).
S@lution: Since./{x) = ax2 +bx+ c has a relative maximum at (5, 12),
therefore /'(5) = 0
f'(x) = 2ax + b
/'(5) = 0 ⇒ 2a(5) + b = 0
⇒ b = - 10 a ... ( 1)
Again./{x) = ax + bx+ c crosses y - axis at (0, 3).
2

So/(0) = 3
⇒ a.0 2 + b.O + c = 3
⇒ c= 3 ... (2)
Also (5, 12) is a point in the curve.
So,/{5) = 12
⇒ 25a + 5b + c = 12
⇒ 25a + 5 (- lOa) + 3 = 12, using (1) and (2)
⇒ 25a=-9

9
⇒ a=--
25

So b = - 10 (--2._J =~
' 25 5

Hence the required function is./(x)= (-J__J

25 X
2
+
(18]
5 x+ 3
eo,,.ca_vity and Inflection Points 3.23

• ( EXERCISE - 3 )

J. Determine the Concavity and inflection points of the following functions:

(a) _f(x) = y' + 3x2 - 9x - 8 (b) Rx) = x2 + 4

.I\ 2x

,J I X
(c) f(x) = 2
~ + x - 1Sx + 3 (d) /{x) = - +-
x 16

In x+ x
(e) f(x ) =xeh· (f) .f{x)= - -
x
1
(g) .f(x) = .r' - 6x + 12x
2. Find all inflection points of the graph of the following functions:
2
(a) .f(x) = J.A -6x + 8x+ 10 (b) .f{x)=x113 + 2

(c) f(x ) =
IO
- +- 2
X
10
X
(d) j(x) ~✓ x -x 4
(e) .f{x) =xex (f) .f{x) = x + tan- 1x

(g) f(x ) = sin2x - 2 cosx, xE[0, 21t] (h) .f{x) = x 113 (27 -x)

(i) .f(x) = x4 - 4x3 + 10 (j) .f{x) = sin x + cos x, x E [0, 21t]

3. Using first derivative test, classify the given critical numbers as relative minimum, relative maxnnu
m
or neither.
e-x2 1
(a) .f(x) - - , at x = 1, - (b) .f{x) =(x3 - 48x) 113 at x = 4
3-2x 2
(c) j(x) = x - 2 sinx, xE[0, 21t]
s
4. Using second derivative test, determine whether each critical number of the following function
corresponds to a relative maximum, a relative minimum or neither.

(b) .f{x) = x2 -x+5

(a) f(_x) = 3x5 - 5x 3 + 2 x+4
5· Determine the intervals of increase and decrease for the following functions:

X
(b) .f{x ) = x2 e-Jx (c) .f{x) = (lru)2
(a) .f{x) = -
x2 +I
of
6. Given the properties of a function f given below. Determine the minimum and maximum values
/{x), inflection points, intervals in which/ is concave up/down.
2
x 2 - 7x 8 x + 8x - 20
f'(x) _x_+__-_ and /" (x) = (x+ ) 2 ,f(-4) is not defined.
4
4
 ~ "' / ,, .,., , . .\:.,(~. :s' '.~"./, , 'y.; ' / .
', t: , ,. ,
.....,, . .. <•. 1i;,; ' . CQl
3.24 . . ~l~
.
. b 1 w Determine the rrummum and maxiJnu ~
7. Given the properties Of a function / given e o .
. hich/ is concave up/down. tn Valu
esof
j(x), inflection points , intervals I~ w == 0 /'(2) ==0,
.f{-2) = - 1,/(2) = - 1,/(0) == 1,/ (- 2) ' 1 <1
f '(0)= 0 f"(x)> Ofor x<-1 and x> 1, f"(x) <Of~ r- < x .
' · functions:
8. Determine the inflection points of the following

x +l (b) xll 3 (27 - x)

(a) j(x) = -
x 2 +I

x 2+2x- 3 (d) .f{x) = x3 + 3x2 - 9x + 2

(c) f(x) = x 2
-3x + 2
(e) f(x) = x-S13 (x2 - 4) (f) .f{x) = sin2x - cosx, x E [- n, 3n]

x4 1
(g) f(x) = 1 + x2 - - (h) .f{x) = -+ 4x2
2 X
(i) f(x) = x3e-4x (j) _f{x) = x 2 ln(x + 2)
9. Find all possible extreme values of each of the following functions by apply
ing the first derivative
test:
(a) .f{x) = x4- 8.x3 + 22.x2 - 24x + 12 for all x ER
(b) .f{x) = (x - 1)2 (x + I) for all x E R
10. Find the range of values of x for which y = .x4 - 6x3 + l 2x2 + 5x + 7
is concave upwards or
downwards. Find also the points of inflection, if any.
11. Find the intervals in which the curve y = (cos x + sin x ) et is concave
upwards or downwards,
when x lies in the interval (0, 2n).

ADDITIONAL EXERC:ISES ~~~~~~

'(C!i'-,,;,';f,Fi'.;E.k,:ih'\ . - - - - - - -
~' .:i';ii(.(i' }/if'Lii, ,,,,· ·, ..
1. ~se th~ first and s~ond derivative off, given by f (x) to determine the
mcreasmg, decreasmg, concave up and concave down. Also find the inflec
intervals on which/is l
3 2
tion point, if any :
(a) /(x)= x - 3x +I (b) /(x)= x +2sin x
(c) f(x)= x2 -3x+ 8 (d) f(x) =x4 - 5x3 +9x2
(e) f(x)= 3x 4 -4x 3
( f) / ( X) == ( 2x + 1 )3
f(x) x-2 2
(g) (X 2
+ X + 1)

2.
(i) /(x)= (x
2 3
' -Ir
Find the intervals on which the followin fi . . . creasing,
decreasing, concave up, concave down
•
A7
u~ctions f 0 _ver the specified interval is 1l1
( ) • 2
(1) f x =sm 2x,xE [0, n]
• so md the pomts of inflection.
(") ( )
u f x = 2x+c otx, x E (0, n)
eoncavity and Inflection Points 3.25

(iii) f ( x) =sin x - cos x, x E ( -1t, 1t) (iv) /(x) = l - tan( x/2) ,xE( -1t, n)

(v) J(x) =(sinx + cos x )' ,x E [-it, it] (vi) f (x) =secx tanx , x e (-; , ; J
3. Examine the validity of the following statements
g
(i) If/a nd g are concave upon an interval, then so if/+
is/g
(ii) If/and g are concave up on or interval, then so
and usin g/, determine whe re/ is concave
4. Find the inflection point of the following poly nom ials/
up and concave down
2 4 3
3
(i) f ( x) = x - 3x + 2x (ii) f(x) =x -4x +10
3 2
(iii) f ( x) =x -5x
5 4
- 240 (iv) f(x) =x -12 x

4 5 2
(v) f ( x ) =sx +16x -25
3 2 tly one inflection point
5. Show that f ( x) = ax + bx + ex + d ( a "# 0) has exac
curve
6. What can you say abou t the inflection points of the
(a) y =ar2 +bx + c, a ;c: 0
(b) y = x 3 + bx2 +ex + d

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