
Applications
of differentiation  the graph of a function and its derivatives 
Finding and classifying
critical (or stationary) points 
Finding
extreme points 
Finding points of
inflection 
Finding and classifying
critical (or stationary) points, examples 






Finding and classifying
critical or stationary points examples 
Example: Find
the extreme point of the quartic polynomial y
= x^{4} 
4x^{3} + 6x^{2} 
4x and sketch its graph. 
Solution: To
rewrite given polynomial from general into translatable form, and this
way determine its source form,
we should always first calculate the coordinates of translation 

then,
plug x_{0}
and y_{0}
into 
y
+ y_{0} = a_{4}(x + x_{0})^{4}
+ a_{3}(x + x_{0})^{3} + a_{2}(x
+ x_{0})^{2} + a_{1}(x +
x_{0}) + a_{0} 
to get the source form of the given polynomial 
y

1 = 1· (x + 1)^{4} 
4(x + 1)^{3} + 6(x + 1)^{2} 
4(x + 1) + 0,
so the source function y
= x^{4}. 
Therefore,
we can write, y
= x^{4} 
4x^{3} + 6x^{2} 
4x or
y +
1 = (x 
1)^{4} or
f(x)
= (x 
1)^{4} 
1. 
To
find the extreme point, calculate the first derivative of the above
expression and set it to zero, 
f
' (x) = 4(x 
1)^{3},
then f
' (x) = 0
or 4(x 
1)^{3} = 0
what yields x
= 1 
as
possible abscissa of the extreme point. 
To
check if x =
1
is the abscissa of the extreme point, plug it into
successive higher order derivatives, starting with
the
second derivative. Thus, 
f
'' (x) = 12(x 
1)^{2},
since f
'' (1) = 0
we must proceed with the higher order derivatives, so 
f
''' (x) = 24(x 
1)
and f ''' (1) =
0
and finally f
^{IV} (x) = 24 > 0 
confirms
x = 1
is the abscissa of the minimum as
first of higher order derivatives that do not vanishes
at this point
is of even order and positive. 
By plugging x =
1
into given quartic we find the
minimum
T_{min}(1,
1),
as
shows
the figure below. 


Example: Find
extreme points and points of inflection of the quintic y
=  x^{5} +
5x^{4} 
7x^{3} + x^{2} +
4x
 1
and sketch its graph. 
Solution: To
rewrite given polynomial from general into translatable form, and this
way determine its source form,
we first calculate the coordinates of translation 

then,
plug x_{0}
and y_{0}
into 
y
+ y_{0} = a_{5}(x + x_{0})^{5}
+ a_{4}(x + x_{0})^{4}
+ a_{3}(x + x_{0})^{3} + a_{2}(x
+ x_{0})^{2} + a_{1}(x +
x_{0}) + a_{0} 
to get the source form of the given polynomial 
y
+ 1 = 
1(x + 1)^{5} + 5(x + 1)^{4} 
7(x + 1)^{3} + (x + 1)^{2} + 4(x + 1)

1, 
what
after expanding and reducing gives y
= 
x^{5} + 3x^{3},^{
}the source quintic. 
Therefore,
we can write
y
= x^{5} +
5x^{4} 
7x^{3} + x^{2} +
4x
1 
or y

1
=  (x 
1)^{5} + 3(x 
1)^{3} or
f (x)
=  (x 
1)^{5} + 3(x 
1)^{3} +
1. 
To
find the extreme points, calculate the first derivative of the above
expression set it to zero and solve for x, 
f
' (x) = 
5(x 
1)^{4} + 9(x 
1)^{2} = 
(x 
1)^{2} · [5(x 
1)^{2} 
9] 
then
we set
f
' (x) = 0,

(x 
1)^{2} · [5(x 
1)^{2} 
9] = 0 
what yields, x_{1}
= 1,
x_{2}
= 1 
3/Ö5
and x_{3}
= 1 + 3/Ö5
as potential abscissas of the extreme points. 
To
check if x_{1}
= 1
is the abscissa of the extreme point, plug it into
successive higher order derivatives, starting with
the
second derivative. Thus, 
f
'' (x) = 
20(x 
1)^{3} + 18(x 
1) = 
2(x 
1) · [10(x 
1)^{2} 
9]
since f
'' (1) = 0 
we must proceed with the higher order derivatives, 
f
''' (x) = 
60(x 
1)^{2} + 18
so that f ''' (1) =
18
is not 0 
what
classifies root x_{1}
= 1
as the abscissa of the point of inflection, as
first of higher order derivatives that do not vanishes
at this point is of odd order. 
While
roots, x_{2}
= 1 
3/Ö5
and x_{3}
= 1 + 3/Ö5
plugged into the second derivative respectively yield 
f
'' (1 
3/Ö5)
> 0,
so x_{2}
= 1 
3/Ö5
is the abscissa
of the minimum

and f
'' (1 + 3/Ö5)
< 0,
so x_{3}
= 1 + 3/Ö5
is the abscissa
of the maximum.

To
find points of inflection of the given quintic we set the second
derivative to zero and solve for x, 
f
'' (x) = 
20(x 
1)^{3} + 18(x 
1),

2(x 
1) · [10(x 
1)^{2} 
9]
= 0 
what yields, x_{1}
= 1,
x_{2}
= 1 
3/Ö10
and x_{3}
= 1 + 3/Ö10 as
possible abscissas of the points of inflection. 
The
root x_{1}
= 1
we've already classified
as the point of inflection. 
Then
we plug roots, x_{2}
= 1 
3/Ö10
and x_{3}
= 1 + 3/Ö10
into the third derivative respectively 
f
''' (x) = 
60(x 
1)^{2} + 18,
so
f ''' (1 
3/Ö10)
¹
0
and
f ''' ( 1 + 3/Ö10)
is not 0 
what
classifies roots, x_{2}
and x_{3
}as the abscissas of the points of
inflection. 
By
plugging the abscissas x_{1},
x_{2} and x_{3}
into the given quintic we get the ordinates of the points of inflection. 
The
graph of the given quintic polynomial and its source function is shown
in the figure below. 









Functions
contents F 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 