CHAPTER 3

APPLICATION OF DERIVATIVES
3-1. Extreme value of a function
Definition

A function f has a maximum value on an interval I, if there is a number d in I such

that f(x) < f(d), for all x in I. We call f(d) the maximum value of f in I. similarly a function f
has a minimum value in I, if there is a number c in I such that f(c) < f(x), for all x in I. We
call f(c) the minimum value of f in I. These which are either maximum or minimum value
are called extreme value of f in I.

Theorem

If a function f is continuous in a closed interval [a, b], then f has maximum and minimum
value in [a, b].

Example
2
Let f ( x )=x−x and [a,b] = [0, 1] . Since f is continuous in [0, 1] f has maximum and
minimum value.

Theorem

Let f be continuous on [a, b]. If an extreme value of f on (a, b) occurs at c in (a, b) , then

f ‘ (c) = 0 or f ’ (c) doesn’t exist. Such number c in (a, b) at which f ‘ (c) = 0 0r f ‘ (c)
doesn’t exist is called critical number.

To find extreme value of f in [a, b] find:

1. Find f(a)

2. Find f(b)

3. Critical number c

Then maximum value of f in [a, b] = max imum { f ( a), f (b ), f ( c ) }

Then minimum value of f in [a, b] = min mum { f ( a), f ( b), f ( c ) }

Example
1
Let f(x) = x – x3. Find all the critical number and the extreme value of f in (0, 1)

Ans

f ' (c )= 0
1−3 c 2 =0
c=±
√3 but −√3 ∉(0 , 1) hence c=√ 3
3 3 3
f (a )=f (0)=0
f (b )=f (1)=0

( ) ( )
3
f (c )=f
√ 3 √3 √3
= − =√ − √ = √
3 3 3 27 3−3 √3 8 √ 3
=
. 3 3 3 3 27 27 9

Maximum value of f in [0, 1] =

{ }
max imum { f ( a), f (b ), f ( c) }= 0 , 0 ,
8 √3 8 √3
9
=
9

min imum { f (a ), f (b ), f (c)}= {0 , 0 , √ }=0

8 3
Minimum value of f in [0, 1] = 9

Relative extreme values

Definition

A function f has a relative maximum value (relative minimum value) at a number c in [a,
b], if there is a number
δ >0 such that f ( c ) is maximum value ( minimum value ) in (c−δ , c+δ ) A value which is
relative maximum value (relative minimum value) is called relative extreme value.

Example
3
Let f ( x )=x −3 x+2 . Find all relative extreme values.

Ans.
2
 f ( x )=x 3 −3 x+2
f ' ( x )=3 x 2 −3
f ' (c )=3 c 2−3
To find critical number f ' (c )=0 ⇒3 c 2 −3=0
⇒ c=±1
Since f(-1)=0 and f(1)=-4. The -4 is the relative minimum value and 0 is the relative
minimum value.

3-2. Rolle’s Theorem, mean value theorem and their applications.

1. Rolle’s theorem.
If f is continuous function on [a, b] and differentiable on (a, b) such that f(a)
= f(b), then there exist c ∈(a , b) such that f ' (c )=0 .

Proof.

If f is constant function, since f ‘ (x) = 0 for all x in (a, b) in particular there

exist c in (a, b) such that f ‘ (c) = 0.If f is not constant function, since f is
continuous in[a, b] f has extreme value in [a, b] by the theorem we see
previously and its maximum or minimum value is distinct. Let one of the
extreme vale be m. Since f(a) = f(b) neither f(a) or f(b) can’t be the
extreme value. So the extreme value must be f(c) wherec ∈(a , b) . Since f
is differentiable in (a, b), c is the critical number such that f ‘(c) = 0.

2. Mean value theorem

If f is continuous function on [a, b] and differentiable on (a, b), then there
f (b )−f (a )
f ' (c )=
exist c ∈(a , b) such that b−a .

Proof

Let us defined a function g on [a, b] by

(
g( x )=f ( x )− f (a )+
f ( b )−f (a )
b−a )
( x−a ) , for x ∈[ a , b ]
. Since f is continuous
on [a, b] and differentiable on (a, b), g is continuous on [a, b] and

3
 differentiable on (a, b) such that g(a) = g(b). By Rrolle’s theorem there
exist c ∈(a , b) such that g ' (c )=0 .

Since
g ' (x )=f ' ( x )− ( f (b )−f (a )
b−a ),

⇒ 0=g ' (c )=f '( c )− ( f (b )−f ( a )

b−a )
, for c ∈(a , b )

⇒ 0=f ' (c )− ( fb−a

(b )−f ( a )
) , for c ∈(a , b )
f (b )−f ( a )
⇒ f ' (c )=
b−a

We can use these theorems to show that If f and g are continuous functions in [a, b] and
differentiable on (a, b) such that f ‘(c) = g’(c) for ∀ c∈(a , b ) , then f – g is constant
function on [a, b]. This can be one of the applications of these theorems.

Proof.

Since f ' (c )=g '(c )⇒ ( f −g ) ' (c )=0 ⋅¿⋅(1 )

Since f and g are continuous on {a, b] and differentiable on (a, b). f – g is continuous
on[a, b] and differentiable on (a, b).

For arbitrary y& z let a<y<z<b. From this one can see that f – g is continuous on[y, z]
and differentiable on (y, z).

By mean value theorem for [y, z], there exist c in (a, b) such that

(f −g )( z )−( f −g )( y )
(f −g ) ' (c )=
z− y
Using (1)

( f −g )( z )−(f −g )( y )
0=
z−x
⇒(f −g)( z )− ( f −g )( y )=0
⇒(f −g)( z )= ( f −g )( y )
Hence f – g has same vale for all x in (a, b). This implies that f – g is
constant function.
4
3-3. Monotonic function

Definitions

1. A unction f is said to be increasing on interval I if and only if

f ( x )≤f ( y ) whenever x < y for all x , y in I .

2. A unction f is said to be strictly increasing on interval I if and only if

f ( x )<f ( y ) , whenever x < y for all x , y in I ..

3. A unction f is said to be decreasing on interval I if and only if

f ( x )≤f ( y ) whenever x > y for all x , y in I .

4. A unction f is said to be strictly decreasing on interval I if and only if

f ( x )<f ( y ) whenever x > y for all x , y in I .

5. A function which is either increasing or decreasing function is

called monotonic function.

Theorem.
Let f be continuous on a closed interval I and differentiable at each interior points of I.

a. If f ' ( x )≥0 at each interior points of I, then f is increasing on I, moreover f is

strictly increasing on I if f ‘ (x)=0 for at most a finite numbers of points in I.

b. If f ' ( x )≤0 at each interior points of I, then f is decreasing on I, moreover f is

strictly decreasing on I if f ‘ (x)=0 for at most a finite numbers of points in I.

Proof

(a) Let y and z be arbitrary numbers in I, with y<z. By the assumption f is

continuous on [y, z] and differentiable on (y, z). By mean value theorem there
f ( z )−f ( y )
c ∈( y , z ) such that f ' (c )= ⇒ f ' (c )( z− y )=f ( z )−f ( y )⋅¿⋅(1 )
exist z− y

We are given that f ' (c )≥0 since z>y and hence z – y >0, therefore we can
say that f ' (c ) ( z− y ) ≥0⋅¿⋅(2 ) . Using (2) in (1)

f ( z )−f ( y )≥0⇒ f ( z )≥f ( y ).

5
 Since y and z are arbitrary this shows that f is increasing function in I

(b) This can be proved similarly ( proof exercises)

Example.
3 2
Let f ( x )=2 x +3 x −12 x−3 . Find intervals such that f is strictly increasing and f
is strictly decreasing.
3 2 2
Since f ( x )=2 x +3 x −12 x−3 ⇒ f ' (x )=6 x +6 x−12=6( x +2)( x−1 )

Hence let us use sign chart for f ‘ (x) to see interval in which f ‘ (x) is positive(+) or

f ‘ (x) is negative(-) and f ‘ (x) =0

-2 1
X+2 - - - - 0 + + + + + + + + + + +

X–1 - - - - - - - - - - 0 + + + + +

F ‘(x)=6(X+2)(x-1) + + o - - - - - - - 0 + + + + +

From the above sign chart

f ' ( x )≥0 for x ∈(−∞ ,−2 ]∪[1 , ∞) and f ' (x )≤0 for x ∈[−2 ,1 ]

Therefore f is strictly increasing for all x ∈(−∞ ,−2 ]∪[ 1 ,∞ ) and f is strictly decreasing
for all x∈[−2,1] .

Exercises

1. Find critical number of the following functions.

1
f ( x )=
2
a. f ( x )=x + 4 x +6 b. √ x 2+ 1
2. Find extreme values of the given function in the given interval.

6
 4
a. f ( x )=x −4 x on [-4, 4] b. f ( x )=√ 1+x 2 on [-2,3]

3. Find the relative extreme value of the following function.

3 2
a. f ( x )=x +3 x −9 x−2 b. f ( x )=−3−Sinx

4. Find a number c in the interval (a, b) for which the line tangent to the graph of f at
(c, f(c))is parallel to the line joining (a, f(a)) and (b,f(b)).
1
3 3
a. f ( x )=x −6 x where (a , b )= (−1 ,3 ) b. f ( x )=1+x where ( a , b)= (1 , 2)

5. Determine the function f satisfying the following.

x x π
f ' ( x )=sec tan and f ( )=2
a. f ' ( x )=−2 and f (0 )=0 b. 2 2 2

6. Find interval in which f is strictly increasing and strictly decreasing.

4 3
a. f ( x )=x −2 x −9 x b. f ( x )=√ 16−x2

3-4. The first and second derivative test

Theorem (First derivative test)

Let f be continuous on a closed interval I, and c be in I.

a. If f ' changes from positive (+) to negative (-) at c, then f has a relative
maximum value at x = c.

b. If f ' changes from negative (-) to positive (+) at c2, then f has a relative
minimum value at x = c2.

7
C2

c
 Proof

a. Since f ‘ changes from positive to negative at x = c, there is a number δ >0 such

that

f ‘(x) >0 for all x ∈(c −δ , c ) and f ‘ (x)<0 for all x ∈(c , c +δ ) . Using theorem on
monotonic function f is strictly increasing on (c−δ , c ) and f is strictly decreasing
on (c , c+δ ) . Therefore f(c) is relative maximum value of f on (c−δ , c+δ ) .

b. Part (b) can be proved similarly.

Example 1.
3 2
If f ( x )=4 x +9 x −12 x +3 , then find all relative extreme value of f.

Ans. Let us used sign chart to solve the problem.

f ( x )=4 x 3 +9 x 2 −12 x +3 ⇒ f ' ( x )=12 x 2 +18 x−12
=6 ( 2 x 2 +3 x−2 ) )
=6 ( x+2 )(2 x−1 )

1
-2 2
X+2 - - - - 0 + + + + + + + + + + +

2X – 1 - - - - - - - - - - 0 + + + + +

Since f ‘ changes form positive to negative at x =-2, by first derivative test

3 2
(−2 )=4(−2) +9(−2
F f‘(x)=6(X+2)(2x-1) + + )o−12(−2)+3=39
- - - - -is- the -relative
0 + maximum
+ + +value
+ of f.

1
x=
Also since f ‘ changes form negative to positive at 2 , by first derivative test
8
 f (−2 )=4 () () ()
1 3
2
+9
1 2
2
1
−12 +3=
2
−1
2 is the relative minimum value of f.
2
Example 2. Draw the graph of f ( x )=( x−1 ) ( x−3 )

f ( x )=( x−1 )2 ( x−3 ) ⇒ f ' ( x )=2( x−1 )( x−3)+( x−1)2

=( x−1)(2 x−6+x −1)
Ans. =( x−1)(3 x−7 )

Let us sign chart to determine:

1. intervals in which f is strictly increasing and strictly decreasing and

2. Relative minimum value and relative maximum value and hence relative
minimum point and relative maximum point.
7
1 3

x-1 - 0 + + +

3x-7 - - - 0 +

(x-1)(3x-7) + 0 - 0 +

1. From the above sign chart:

7
x ∈(−∞ , 1]∪[ , ∞ )
a. f is strictly increasing for 3 and strictly decreasing for all
7
x ∈[ 1 , ]
3
2
b. f (1)= (1−1 ) ( 1−3 ) =0 is the relative maximum value and

( ) ( )( )
2
7 7 7 5
f = −1 −3 =−1
3 3 3 27 is the relative minimum value. i.e (1,0) is

(
relative maximum point and 3
7
, −1 )
5
27 is relative minimum point.
9
 2. y- intercept of the graph is -3 and x – intercept of the graph are 1 and 3

3. As we go further to the left the graph open down wards and as we go further to
the right the graph open upwards.

4. At x= 1 the graph touch the x - axis and return back and at x= 3 the graph cross
the

X – axis.

-3 -2 -1 0 1 2 3 4

-1
.

-2

-3

Theorem (Second derivative test)

Assume that f ‘ (c) = 0

1. If f ‘’(c) <0, then f(c) is relative maximum value of f

2. If f ‘’(c) > 0, then f(c) is relative minimum value of f

10
 3. If f ‘’(c) = 0, then from this test alone we can’t draw any conclusion about relative
extreme value of f.

Proof.

f '' ( c )= lim f ' ( xx−c

)−f ' ( c )
<0
(1.) x →c

Since f ’ (c) =0, for

f ' ( x )−f ' (c ) f ' (x )
x ∈(c −δ , c +δ ) and x≠c , = <0 ⋅¿⋅(∗)
x−c x−c

If c−δ< x <c , then x−c <0⋅¿⋅(** ) . Using (*) and (**) f ' x )>0

If c < x <c +δ , then x−c> 0⋅¿⋅( ***) . Using (*) and (***) f ' x )<0

Hence f ‘ changes form positive to negative at x = c. Using first derivative test

f(c) is relative maximum value of f.

(2). The proof of (2) is similar as (1) and it is left as exercises.

Example.
3
Let f ( x )=x −3 x . Using second derivative test find the relative extreme values of
f.

Ans.

f ( x)=x 3 −3 x⇒ f ' (x )=3 x 2−3=3( x−1 )(x+1 )

f ' ( x)=0 ⇒3( x−1 )( x+1)=0
⇒ x=−1 or x=1
Hence c=−1 or c=1 are critical numbers
f '' ( x )=6 x
3
Since f '' (−1)=6(−1)=−6<0 ⇒ f (−1)=(−1 ) −3(−1)=−1+3=2 is relative maximum
value.
3
And since f '' (1)=6(1 )=6>0 ⇒ f (1 )=(1 ) −3(1)=1−3=−2 is relative minimum value

11
3.5 Application to extreme values and related rates.
The method of finding extreme value of a function can be applied to solve
practical optimization problems. It can be applied to solve problem of finding
maximum area, volume, profit, etc and it can be applied to find minimum
distance, time and cost etc.

The method of solving such problem will depend upon appropriate functional
representation of the problem on a closed interval. B/c if a function is
continuous on a closed interval it has maximum and minimum value.

Steps to solve practical problem

1. Understand the problem

2. Draw appropriate diagram.

3. Assign variables to the quantity that is going to be maximized or

minimized.

4. Express the quantity to be maximized or minimized in terms of another

variable.

5. Use methods of finding extreme vales to the function expressed in step

4.

Example.

A metal box without top cover is wanted to be constructed from a square metal sheet of
sided length 10m. When it is constructed first a square pieces of the same size from
each corner is taken out and then by folding up each side’s the metal box is
constructed. Find the dimension of the taken out metal to get maximum volume of the
box.

Ans Let x be the length of the taken out metal.

x
x

10-2x

10-2x
From the second graph volume v can be found.
x 10 – 2x
v (x )=x ( 10−2 x )2 x∈[ 0 , 5 ]
X 10 – 2x for x
12
Let us find the maximum value of v

v (x )=x (10−2 x )2 =4 x 3 −40 x 2 +100 x ⇒ v ' ( x )=12 x 2 −80 x+100=4 (3 x−5 )(x −5)
5
v ' ( x )=0 ⇒ 4(3 x−5 )( x−5)=0 ⇒ x= or x=5
3
5
The critical numbers are 3 or 5.

v '' (x )=24 x−80

Since v '' (5 )=24 (5 )−80=120−80=40>0⇒ v(5) is the minimum value

But
v '' () ()
5
3
=24
5
3
−80=40−80=−40 <0 ⇒ v
5 5
3 3
5 2 2000
= 10− =
3 () (
27 is the maximum )
value.

5
Hence the dimension of the taken out metal is 3 to get maximum volume
2000
cub .unit
27

Related Rates
Derivative can be used to find related rates of change.

Example

Velocity .

The average velocity of an object travels during time t is

dis tan ce traveled s f −si

V= =
time changed t f −t i

If f represent distance which is a function of time

dis tan ce traveled f (t f )−f (t i )

V= =
time changed t f −t i

lim f ( t t)−f
−t
(t )
0
0
=f ' ( t )
0
Instantaneous velocity at t0 = t →t 0

13
When we relate two rate of change with respect to the same variable we say we have
related rates.

For example when a spherical balloon is inflated, the radius r and the volume v of the
balloon are changed as a function of time t.

4
V = π r3
3

Differentiating both sides with respect of time

dV 4 dr dV dr
= π ( 3 r2 ) ⇒ =4 πr 2 ⋅¿⋅(∗)
dt 3 dt dt dt

dv dr
and
Hence dt dt are related by (*). And we have related rates.

dv dr
, , and r
From (*) if we known two of the quantity from dt dt using the equality we can
find the other quantity.

Example .

A spherical balloon is inflated at the rate of 10 cub . /min. Find the rate of the radius of
the balloon changed per min, when the radius of the water is 5cm.

Ans.

dV dr
=4 πr 2
dt dt
dr
⇒10=4 π (5)2
dt
dr 1
⇒ = cm /min
dt 10 π

3.6 Concavity and inflection points

Definitions.

14
1. Let f be differentiable at x=c , and the line lc be the tangent line to the graph of
f at (c, f(c)). The graph of f is concave up ward at (c, f(c)) if there is an open
interval Ic about c such that if x is in Ic and x≠c , then (x, f(x)) lies above the
line Ic .

2. Let f be differentiable at x=c , and the line lc be the tangent line to the graph of
f at (c, f(c)). The graph of f is concave down ward at (c, f(c)) if there is an
open interval Ic about c such that if x is in Ic and x≠c , then (x, f(x)) lies below
the line Ic .

3. The graph of f is concave upward on an interval I, if it is concave upward for

all (x, f(x)) where x ∈ I . And the graph of f is concave upward on an interval I,
if it is concave upward for all (x, f(x)) where x ∈ I .

x1 x2

From the above figure

a. for all x ∈(−∞ , x1 ) the graph concave down ward

b. for all x ∈( x 2 , ∞) the graph concave down ward

15
 c. for all x ∈( x 1 , x 2 ) the graph concave up ward

Theorem

a. If f ‘’ (x) >0 for all x in an interval I, then the graph of f concave upward

b. If f ‘’ (x)<0 for all x in an interval I, then the graph of f concave down ward

Inflection points

Definition

Points at which the concavity down ward is changed to concavity upward or

Points at which the concavity up ward is changed to concavity down ward is
called inflection points.

Form the above figure ( x 1 , f ( x 1 ) and ( x 2 , f ( x 2 ) are inflection points.

Example
4 2
Find all inflection points of f ( x )=x −6 x + 8 x+10

Ans.
f ( x )=x 4 −6 x 2 +8 x+10 ⇒ f ' ( x )=4 x 3−12 x +8 ⇒ f '' ( x )=12 x 2−12=12(x −1)( x +1)
Let us use sign chart to find inflection points.
-1 1

x-1 - - - 0 +
x+1 - 0 + + +

12(x-1)(x+1) + 0 - 0 +

From the above sign chart

1. (-1,f(-1)) and (1, f(1)) are inflection points.

2. The graph of f concave upward for all x∈(−∞ ,−1 ) and for all x ∈(1 , ∞)

16
 3. The graph of f concave down ward for all x∈(−1 ,1 )

3.7 Graphing ( Sketching)

Graph of the function is the set of points represented all coordinate that satisfies the .
/function. All the concepts discussed so far are used to graph a function. Let us list by
the following table.

property Method of finding the property

Let x = 0 and find corresponding

value
f has y – intercept at x = c
for y. This y value is y - intercept

Let y= 0 and find corresponding

value
f has x – intercept at x = c
For x. This x value is x - intercept

f(c) is relative maximum value of f

f ‘ (c)=0 and f ‘ change form + to –
Or f ‘ (c)=0 and f ‘’ (c)<0

f(c) is relative minimum value of f

f ‘ (c)=0 and f ‘ change form –to +
Or f ‘ (c)=0 and f ‘’ (c)>0

f is strictly increasing on an interval I f ‘ (x) >0 for all x in I

f is strictly decreasing on an interval I f ‘ (x) >0 for all x in I

Graph of f concave upward on interval I f ‘’(x) > 0for all x in I

Graph of f concave down ward on interval I f ‘’(x) < 0for all x in I

(c, f(c)) is inflection point of the graph f ‘’ (c)=0 and f ‘’ changes in sign.

17
3.8. Tangent line approximation and differentials
To describe a general procedure for approximating value of a function by means of
calculus , let f be differentiable at x = a. By the definition of f ‘ (a) if
f ( x )−f (a )
x → a , then → f ( a)
x−a . Hence for values of close enough to a , f(x) – f(a)
close to f ‘ (a)(x – a). To emphasize that only values of x close to a are to be
considered. We usually replace x by x+h, and obtaining

f (a+h )≈f ( a)+ f ' (a )h (∗)

Notice that x is close to a iff h is close to 0. (*) is called tangent line approximation. The
approximation in (*) is most useful when f(a) and f ‘ (a) are easy to compute.

Example. Use tangent line approximation to estimate the value of √ 8 .5

Ans. Let f ( x )=√ x and a = 9 and h = -0.5. We seek f (8 .5 ) .

1
f ' ( x )=
2√x .

1
f (a+h )≈f ( a)+f ' (a )h ⇒ f (8 .5 )≈f (9 )+f ' (9 )(−0. 5 )⇒ f (8 . 5)≈3+ (−0 . 5)=2. 91667
6

18