LESSON 8 DERIVATIVES OF POLYNOMIAL FUNCTIONS
Theorem If f and g are two differentiable functions and k ( x )  f ( x )  g ( x ) ,
then k ( x )  f ( x )  g ( x ) . That is, the derivative of a sum is the sum of the
derivatives.
Proof By definition, k ( x )  hlim
0
k( x  h )  k( x )
.
h
k( x  h )  f ( x  h )  g( x  h )
k( x )  f ( x )  g( x )
k( x  h )  k( x ) = f ( x  h )  g( x  h )  [ f ( x )  g( x ) ] =
f ( x  h )  f ( x )  g( x  h )  g( x )
k( x  h )  k( x )
f ( x  h )  f ( x )  g( x  h )  g( x )
=
=
h
h
f ( x  h)  f ( x)
g( x  h )  g( x )
+
h
h
Thus, k ( x )  hlim
0
k( x  h )  k( x )
=
h
f ( x  h )  g( x ) 
 f ( x  h )  k( x )
lim 

=
h0
h
h


lim
h0
f ( x  h)  f ( x)
g( x  h )  g( x )
+ hlim
= f ( x )  g ( x )
0
h
h
Theorem If f is a differentiable function and k ( x )  c f ( x ) , where c is a
constant, then k ( x )  c f ( x )
Copyrighted by James D. Anderson, The University of Toledo
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Proof By definition, k ( x )  hlim
0
k ( x )  lim
h0
lim
h0
k( x  h )  k( x )
.
h
k( x  h )  k( x )
c f ( x  h)  c f ( x)
lim
= h0
=
h
h
c f ( x  h)  f ( x) 
f ( x  h)  f ( x)
= c hlim
= c f ( x )
0
h
h
Theorem If f and g are two differentiable functions and k ( x )  f ( x )  g ( x ) ,
then k ( x )  f ( x )  g ( x ) . That is, the derivative of a difference is the
difference of the derivatives.
Proof k ( x )  f ( x )  g ( x ) = f ( x )    g ( x )  . By the first theorem above, we

have that k ( x )  f ( x )    g ( x )  . By the second theorem above, we have that
  g ( x )  =   1  g ( x )  =   g ( x )  . Thus, k ( x )  f ( x )  g ( x ) .
Notation: Dx ( f ( x ) )  f ( x ) .
4
3
2
Example 1 If f ( x )  x  5x  3x  x  8 , then find f ( x ) .
Using the properties of derivatives given above, we obtain the following:
f ( x ) = Dx ( f ( x ) ) = D x ( x 4  5 x 3  3 x 2  x  8 ) =
Dx ( x 4 )  Dx ( 5 x 3 )  Dx ( 3 x 2 )  Dx ( x )  Dx ( 8 ) =
Dx ( x 4 )  5 Dx ( x 3 )  3 Dx ( x 2 )  Dx ( x )  Dx ( 8 )
Thus, to find the derivative of this polynomial, we need to know how to
differentiate a variable raised to a positive integer and how to differentiate a
constant.
Copyrighted by James D. Anderson, The University of Toledo
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Consider the following two theorems.
Theorem If f ( x )  c , where c is a constant, then f ( x )  0 .
Proof By definition, f ( x )  hlim
0
f ( x )  lim
h0
f ( x  h)  f ( x)
.
h
f ( x  h)  f ( x)
cc
0
lim
00
= hlim
=
= hlim
0
h0 h
0
h
h
n
Theorem (The Power Rule) If f ( x )  x , where n is a rational number, then
f ( x )  n x n  1 .
Proof We will only prove this theorem for the case where n is a positive integer.
The other cases will be proved in later lessons. By definition, f ( x ) =
f ( x  h)  f ( x)
( x  h )n  xn
lim
= hlim
.
h0
0
h
h
Since n is a positive integer, then we can use the Binomial Expansion Theorem to
n
expand out ( x  h ) . Recall the Binomial Expansion Theorem from Lesson 3:
Binomial Expansion Theorem Let n be a positive integer. Let a and b be real
n
n n  k k
n
n!
n


(
a

b
)

a
b




numbers, then
, where  k 
 
k !( n  k ) ! .
k  0 k
 
( x  h ) n  x n  nx n  1 h 
n ( n  1) n  2 2
n ( n  1) 2 n  2
x
h 
x h
 nxhn  1  h n
2
2
( x  h ) n  x n  nx n  1h 
n ( n  1) n  2 2
n ( n  1) 2 n  2
x
h 
x h
 nxhn  1  h n
2
2
n ( n  1) 2 n  3
 n  1 n ( n  1) n  2


x
h
x h
 nxhn  2  h n  1 
= h  nx
2
2


Copyrighted by James D. Anderson, The University of Toledo
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n ( n  1) n  2
n ( n  1) 2 n  3
( x  h )n  xn
n 1

x
h
x h
 nxhn  2  h n  1
= nx
2
2
h
f ( x  h)  f ( x)
( x  h )n  xn
Thus, f ( x ) = hlim
= hlim
=
0
0
h
h
n ( n  1) n  2
n ( n  1) 2 n  3


lim  nx n  1 
x h
x h
 nxhn  2  h n  1  =
h0
2
2


n xn  1  0      0  0  0  n xn  1
Example 2
If f ( x )  x , then
f ( x )  1
2
If f ( x )  x , then f ( x )  2 x
2
3
If f ( x )  x , then f ( x )  3 x
4
3
If f ( x )  x , then f ( x )  4 x
10
9
If f ( x )  x , then f ( x )  10 x
114
115
If f ( x )  x , then f ( x )  115 x
Returning to Example 1, we have that
f ( x ) = Dx ( x 4 )  5 Dx ( x 3 )  3 Dx ( x 2 )  Dx ( x )  Dx ( 8 )
3
2
3
2
= 4 x  5 ( 3x )  3 ( 2 x )  1  0 = 4 x  15 x  6 x  1
Examples Differentiate the following functions.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
NOTE: The verb differentiate means to find the derivative of.
COMMENT: Don’t forget to identity the derivative before you start the work to
find it.
1.
f ( x )  x2  5
Answer: f ( x )  2 x
2.
f ( x )  3x 2  7 x
Answer: f ( x )  6 x  7
3.
g( x )  x3
2
Answer: g ( x )  3x
4.
f ( x )  x 3  4 x 2  12 x  28
2
Answer: f ( x )  3x  8 x  12
5.
s( t ) 
6t
NOTE: The function s is not a polynomial. However, we can still find the
derivative of this function using the Power Rule. We will have to first
rewrite the expression 6t using properties of radicals before we will be
able to apply the Power Rule. Also, recall from algebra
Since
6t 
6
t 
6 t 1/ 2 , then s( t ) 
1/ 2
NOTE: The Power Rule tells us that Dt ( t ) 
s ( t ) 
n
am  a m/n .
6 t 1/ 2 .
1  1/ 2
t
. Thus,
2
6
6  1/ 2
6
t
= 1/ 2 =
2 t
2
2t
Copyrighted by James D. Anderson, The University of Toledo
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Answer: s ( t ) 
6
2 t
6
NOTE: The expression
2 t
is equivalent to the expression in our answer
for this problem in Lesson 7, which was
3
.
6t
COMMENT: The only time that you should use a slash ( / ) to write a
fraction is for an exponent like we had above. If you anywhere else, you may
1/ 2
be writing something that you don’t want. For example, when the t
was
1  1/ 2
1
 1 / 2 . If you would
differentiated above, the answer was written t
2
2t
1
t 1/ 2
1
 1/ 2

 1/ 2 .
have write 1 / 2 t
, then this expression means
 1/ 2
2
2t
2t
6.
y  3x  20
7.
y  7  2x3  4x5
Answer: y   3
dy
  6 x 2  20 x 4 =  2 x 2 ( 3  10 x 2 )
dx
dy
dy
2
4


6
x

20
x
  2 x 2 ( 3  10 x 2 )
Answer:
or
dx
dx
8.
y
3 2 6
t ( t  6t 3  5 )
2
Copyrighted by James D. Anderson, The University of Toledo
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y
3 2 6
3 8
t ( t  6t 3  5 ) =
( t  6t 5  5t 2 )
2
2
3
3
dy

( 8t 7  30t 4  10t ) =
 2t ( 4t 6  15t 3  5 ) =
dt
2
2
3 t ( 4t 6  15t 3  5 )
3
dy
dy
7
4


(
8
t

30
t

10
t
)
Answer:
or
dt
dt
2
9.
3 t ( 4t 6  15t 3  5 )
h( w )  ( 3w  2 ) ( 4  w )
h( w )  12w  3w2  8  2w = 8  10w  3w 2
h( w )  10  6w
Answer: h( w )  10  6w or h ( w )  2 ( 5  3w )
10.
g( x )  ( 4x  3) 2
g ( x )  16 x 2  24 x  9  g ( x )  32 x  24 = 8 ( 4 x  3 )
Answer: g ( x )  32 x  24 or
g ( x )  8 ( 4 x  3 )
1 4 1 2
5
t  t  4t 
is a position function, then find the
2
3
2
instantaneous velocity function v.
Example
If s ( t ) 
v( t )  s ( t )  2t 3 
1
2
1
t  4 = ( 6t 3  2t  12 ) =  2 ( 3t 3  t  6 ) =
3
3
3
Copyrighted by James D. Anderson, The University of Toledo
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2 3
( 3t  t  6 )
3
Answer: v( t ) 
2 3
( 3t  t  6 )
3
2
Example If s( t )  4t  18t is the position function which gives the position (in
meters) of a particle at time t (in hours), then find
a.
b.
c.
the instantaneous velocity function,
the position and velocity of the particle when t = 0, 2, 3, 6, 7, and 100 hours,
the position(s) when the particle is stopped.
Recall: This problem was worked in Lesson 6. The solutions for Part b and Part c
are the same as what was given in this lesson. The only thing, that is changing, is
Part a. As promised in this lesson, you can write the velocity down in the
following way:
v( t )  s ( t )  8t  18  2 ( 4t  9 )
Example If a ball is thrown upward with a velocity of 36 feet/second from the top
of a 90-foot building, then the height of the ball above ground at time t (in seconds)
2
is given by the position function s( t )  90  36t  16t , where t  0 . Find
a.
b.
c.
the instantaneous velocity function,
the maximum height reached by the ball,
the velocity of the ball when it strikes the ground.
Recall: This problem was worked in Lesson 6. The solutions for Part b and Part c
are the same as what was given in this lesson. The only thing, that is changing, is
Part a. As promised in this lesson, you can write the velocity down in the
following way:
v( t )  s ( t )  36  32t  4 ( 9  4t )
Example Find the equation (in point-slope form) of the tangent line to the graph of
5
4
3
the function y  x  2 x  x  7 x  12 at the point for which x   2 .
Copyrighted by James D. Anderson, The University of Toledo
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Tangent Point: (  2 , y (  2 ) )  (  2 ,  18 )
y (  2 )   32  32  8  14  12   18
y   5 x 4  8 x 3  3x 2  7
m tan  y (  2 )  80  64  12  7  11
Answer: y  18  11( x  2 )
Definition The line perpendicular to the tangent line to the graph of a function is
called the normal line.
 y  f ( x)
Normal Line 
Tangent Point = Normal Point
Tangent Line 
Recall: If two lines are perpendicular, then the product of their slopes is equal to
negative one. If the slope of a line is known, then the slope of the perpendicular
line is the negative reciprocal of this known slope. Since we can find the slope of
the tangent line using the derivative of the function, then the slope of the normal
line is the negative reciprocal of this slope.
Copyrighted by James D. Anderson, The University of Toledo
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Example Find the equation (in point-slope form) of the normal line to the graph of
4
3
2
the function y   x  2 x  x  8 x  35 at the point for which x  3 .
Normal Point: ( 3 , y ( 3 ) )  ( 3 ,  25 )
y ( 3 )   81  54  9  24  35   25
y   4x 3  6 x 2  2x  8
m tan  y ( 3 )   108  54  6  8   68  m normal  
Answer: y  25 
1
1

m tan 68
1
( x  3)
68
3
2
Example Find the point(s) on the graph of y  x  x  8 x  7 at which the
tangent line is horizontal.
The slope of a horizontal line is zero. Since the derivative of the function y will
give us the slope of tangent lines to the graph of y, then we want to find when the
derivative of the function y is zero. That is, we want to solve the equation y   0 .
y  x 3  x 2  8x  7 
y  0 
y   3x 2  2 x  8
3x 2  2 x  8  0 
( x  2 ) ( 3x  4 )  0  x  2 , x  
4
3
4
are the x-coordinates of the points where the tangent
3
3
2
line is horizontal. Now, we need to use the function y  x  x  8 x  7 to find
the y-coordinate of these points.
NOTE: x  2 and x  
x2:
x
4
:
3
y ( 2 )  8  4  16  7   5
365
64
48
288
189
64
16
32
 4



y    


 7 = 
=
27
27
9
3
27
27
27
27
 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
 4 365 

(
2
,

5
)
Answer:
; ,
 3 27 
 y  x 3  x 2  8x  7
Example Determine if there are any tangent lines to the graph of the function
f ( x )  4 x 2  24 x which pass through the point ( 5 ,  42 ) . Write the equation
(in point-slope form) of any such tangent line(s).
First, note that the given point of ( 5 ,  42 ) is not on the graph of
y  f ( x )  4 x 2  24 x  4 x ( x  6 ) since  42  f ( 5 )  4 ( 5 ) (  1)   20 . If
there is a tangent line to the graph of the given function passing through the given
point, then it might look the following:
 y  f ( x )  4 x 2  24 x  4 x ( x  6 )

6
(a, f(a))
( 5 ,  42 )
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Let ( a , f ( a ) ) be a point on the graph of y  f ( x ) whose tangent line passes
through the point ( 5 ,  42 ) .
Then, by algebra, the slope of the tangent line is given by m tan 
f ( a )  (  42 )
a5
since the two points of ( 5 ,  42 ) and ( a , f ( a ) ) are on the tangent line. Since
4a ( a  6 )  42
f ( a )  4a ( a  6 ) , then m tan 
.
a5
By calculus, the slope of the tangent line at the point ( a , f ( a ) ) on the graph of
2
y  f ( x ) is given by m tan  f ( a ) .
Since f ( x )  4 x  24 x , then
f ( x )  8 x  24  8 ( x  3 ) . Thus, m tan  f ( a )  8 ( a  3 ) .
Since these two expressions give the slope of the same tangent line, then they must
be equal. Thus,
4a ( a  6 )  42
 8( a  3)
a5
If this equation has a solution, then there will be a tangent line to the graph of the
given function y  f ( x )  4 x ( x  6 ) that passes through the given point
( 5 ,  42 ) . NOTE: If there is a solution to this equation, it can not be 5 because
this will lead to division by zero on the left-hand side of the equation. Let’s solve
this equation. First, let’s multiply both sides of the equation by a  5 , obtaining
4a ( a  6 )  42  8 ( a  3 ) ( a  5 )
Factoring out the common 2 on the left-hand side of the equation, we obtain
2 [ 2a ( a  6 )  21]  8 ( a  3 ) ( a  5 ) . Now, dividing both side of this equation
by 2, we obtain 2a ( a  6 )  21  4 ( a  3 ) ( a  5 ) . Simplifying both sides of
2
2
the equation, we obtain 2a  12a  21  4 ( a  8a  15 ) .
2a 2  12a  21  4 ( a 2  8a  15 )  2a 2  12a  21  4a 2  32a  60 
0  2a 2  20a  39 . Since the quadratic on the right-hand side of the equation
does not factor, then we will need to use the quadratic formula to solve for a:
Copyrighted by James D. Anderson, The University of Toledo
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a
20 
20  20  4 ( 2 ) ( 39 )
=
4
20 
4  2[ 5  10  1(1) ( 39 ) ]
=
4
20  2 22
10  22
20  2 2[ 50  39 ]
=
=
4
4
2
These two numbers are the x-coordinates of the points on the graph of the given
function y  f ( x )  4 x ( x  6 ) whose tangent line passes through the given point
( 5 ,  42 ) . In order to find the equation of these two tangent lines, we need the ycoordinate of these two points, which we will find using the function
y  f ( x )  4 x ( x  6 ) , and we need the slope of these two tangent lines, which
we will find using the derivative function f ( x )  8 x  24  8 ( x  3 ) .
For a 
10 
22
2
: To find the y-coordinate, we need to find the value of
 10  22 
 10  22 
 f

y



 . Since y  f ( x )  4 x ( x  6 ) , then we have
2
2




 10  22 
 10  22 
 10  22   10  22 12 

  f
  4


y

that 






2
2
2
2
2 







= 2 10 
  2  22 

22 

 = 10 
2




22
  2 
22
=
 20  10 22  2 22  22 = 2  8 22 = 2 (1  4 22 )
 10  22


.
,
2
(
1

4
22
)
Thus, the tangent point is 

2


Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
To find the slope of the tangent line that passes through this point, we need to find
 10  22 
 . Since f ( x )  8 x  24  8 ( x  3 ) , then

f
the value of 

2


 10  22 
 10  22
 4  22 
6 




 = 4 ( 4  22 )

m tan  f
 8

8
=






2
2
2
2





Thus, the equation of the tangent line to the graph of the function

10 
y  f ( x )  4 x ( x  6 ) at the point 

y  2 (1  4 22 )  4 ( 4 
For a 
10 
22
2
22
2

, 2 (1  4 22 )  is given by



10  22 

22 )  x 


2


: To find the y-coordinate, we need to find the value of
 10  22 
 10  22 
 f

y



 . Since y  f ( x )  4 x ( x  6 ) , then we have
2
2




 10  22 
 10  22 
 10  22   10  22 12 

  f
  4


y

that 






2
2
2
2
2 







= 2 10 
  2  22 

22 

 = 10 
2




22
  2 
22
=
 20  10 22  2 22  22 = 2  8 22 = 2 (1  4 22 )
 10  22


.
,
2
(
1

4
22
)
Thus, the tangent point is 

2


Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
To find the slope of the tangent line that passes through this point, we need to find
 10  22 
 . Since f ( x )  8 x  24  8 ( x  3 ) , then

f
the value of 

2


 10  22 
 10  22
 4  22 
6
  8
 = 4 ( 4  22 )
m tan  f 
  = 8






2
2
2
2





Thus, the equation of the tangent line to the graph of the function
 10  22


 is given by
,
2
(
1

4
22
)
y  f ( x )  4 x ( x  6 ) at the point 

2


y  2 (1  4 22 )  4 ( 4 
Answer:

10  22 

22 )  x 


2


Line 1:
y  2 (1  4 22 )  4 ( 4 
Line 2:
y  2 (1  4 22 )  4 ( 4 

10  22 

22 )  x 


2



10  22 

22 )  x 


2


A picture of the two tangent lines:
y  f ( x )  4 x 2  24 x  4 x ( x  6 ) 
Line 2 
 Line 1

6
( 5 ,  42 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
From Lesson 7, the definition of the derivative of the function f at x  a is
f (a  h)  f (a)
defined by f ( a )  hlim
, provided the limit exists. Let
0
h
x  a  h . Then h  x  a . Since x  a  h , then as h  0 , we have that
f ( x)  f (a)
f (a  h)  f (a)
lim
x  a . Thus, f ( a )  hlim
=
. We will
xa
0
xa
h
use this later limit to find f ( a ) in the following examples.
Examples Differentiate the following piecewise functions.
1.
 x 3  64 , x  4
f ( x)   3
 x  27 , x  4
 3x 2 , x  4
f ( x )   2
 3x , x  4
NOTE: Since 4 is a breakup point of the piecewise function, then in order to
calculate the derivative of the function at x  4 , we must use the definition
f ( x)  f (4)

f
(
4
)

lim
of derivative. Thus, we have that
. Since
x4
x4
f ( x )  37
f ( 4 )  64  27  37 , then we have that f ( 4 )  xlim
4
x  4 . Now,
f ( x )  37
lim
in order to find x  4 x  4
we will have to calculate the one-sided
f ( x )  37
f ( x )  37

f
(
4
)

lim

limits f  ( 4 )  x lim
and
 4
x  4
x4
x4 .
f  ( 4 )  lim 
x4
f ( x )  37
x 3  27  37
x 3  64
lim
= x lim
 4
x  4 = x  4
x4
x4 =
x  4   x  4  f ( x )  x 3  27
Copyrighted by James D. Anderson, The University of Toledo
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( x  4 ) ( x 2  4 x  16 )
lim 
lim  ( x 2  4 x  16 ) = 16  16  16 = 48
=
x4
x4
x4
f  ( 4 )  lim 
x4
f ( x )  37
x 3  64  37
x 3  27
= x lim
= x lim
=
 4
 4
x4
x4
x4
x  4   x  4  f ( x )  x 3  64
91
 DNE
0
Thus, f ( 4 )  xlim
4
Thus, the function
undefined.
f ( x)  f (4)
f ( x )  37
lim
=
x4
x4
x  4 = DNE.
f
is not differentiable at x  4 .
Thus, f ( 4 ) is
 3x 2 , x  4
Answer: f ( x )   2
 3x , x  4
NOTE: At the end of this lesson, we will state a theorem which would have
allowed us to determine that the function f is not differentiable at x  4 in a
faster manner.
2.
 2 x 3  18 x 2  25 , x   2

g ( x )   x 4  3x 2  4 x  5 ,  2  x  3
 x 2  11x  115 , x  3

 6 x 2  36 x , x   2

g ( x )   4 x 3  6 x  4 ,  2  x  3
 2 x  11
, x3

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
NOTE: Since  2 and 3 are the breakup points of the piecewise function,
then in order to calculate the derivative of the function at x   2 and x  3 ,
we must use the definition of derivative.
To find g (  2 ) , we must calculate g (  2 )  x lim
2
g( x )  g(  2 )
x  (  2) .
Since g (  2 )   16  72  25  31 , then we have that
g ( x )  31
g ( x )  31
lim
g (  2 )  lim
x2
x  2 . Now, in order to find x   2 x  2 we
g ( x )  31

g
(

2
)

lim
will have to calculate the one-sided limits 
x   2
x  2 and
g ( x )  31
g  (  2 )  lim 
x2
x2 .
Calculating these one-sided limits:
x   2   x   2  g ( x )  x 4  3x 2  4 x  5 . Thus,
g  (  2 ) 
lim
x   2
lim
x   2
g ( x )  31
x2 =
lim
x   2
x 4  3x 2  4 x  5  31
=
x2
0
x 4  3x 2  4 x  36 16  12  8  36

 I.F.
=
22
0
x2
4
2
By the Factor Theorem, x  2 is a factor of x  3x  4 x  36 . The
4
2
factorization of x  3x  4 x  36 can be found using synthetic division.
4
2
4
3
2
Since x  3x  4 x  36 = x  0 x  3x  4 x  36 , then we write the
following in order to carry out the synthetic division
1
0
2
3
4
4
 14
 36
36
1
2
7
 18
0
Copyrighted by James D. Anderson, The University of Toledo
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2
4
2
3
2
Thus, x  3x  4 x  36 = ( x  2 ) ( x  2 x  7 x  18 ) . Thus,
evaluating the given limit, we have that
g  (  2 ) 
=
lim
x   2
lim
x   2
x 4  3x 2  4 x  36
=
x2
lim
x   2
( x  2 ) ( x 3  2 x 2  7 x  18 )
x2
( x 3  2 x 2  7 x  18 )   8  8  14  18   48
x   2   x   2  g ( x )  2 x 3  18x 2  25 . Thus,
g  (  2 ) 
lim
x   2
lim
x   2
g ( x )  31
x2 =
lim
x   2
2 x 3  18 x 2  25  31
=
x2
0
2 x 3  18 x 2  56  16  72  56

 I.F.
=
22
0
x2
3
2
By the Factor Theorem, x  2 is a factor of 2 x  18 x  56 =
2 ( x 3  9 x 2  28 ) . The factorization of x 3  9 x 2  28 can be found using
3
2
3
2
synthetic division. Since x  9 x  28 = x  9 x  0 x  28 , then we
write the following in order to carry out the synthetic division
1
9
2
0
 14
 28
28
1
7
 14
0
2
3
2
2
Thus, x  9 x  28 = ( x  2 ) ( x  7 x  14 ) . Thus,
evaluating the given limit, we have that
g  (  2 )  lim
x   2
=
lim
x2

2 x 3  18 x 2  56
=
x2
lim
x   2
2( x  2 ) ( x 2  7 x  14 )
x2
2( x 2  7 x  14 )  2 ( 4  14  14 )  2 (  24 )   48
Copyrighted by James D. Anderson, The University of Toledo
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g( x )  g(  2 )
=
x2
g is differentiable at x   2 .
Thus, g (  2 )  x lim
2
lim
x2
To find g ( 3 ) , we must calculate g ( 3 )  xlim
3
g ( x )  31
  48 . Thus,
x2
g( x )  g( 3)
.
x3
Since g ( 3 )  9  33  115  91 , then we have that
g ( x )  91
g ( x )  91
g ( 3 )  lim
g ( 3 )  lim
.
Now,
in
order
to
find
x3
x3
x3
x3
g ( x )  91
we will have to calculate the one-sided limits g  ( 3 )  x lim
 3
x  3 and
g ( x )  91
g  ( 3 )  lim 
x3
x3 .
Calculating these one-sided limits:
x  3  x  3  g ( x )  x 2  11x  115 . Thus,
g ( x )  91
x 2  11x  115  91
g  ( 3 )  lim 
lim
=
x3
x  3 = x  3
x3
( x  3) ( x  8)
x 2  11x  24
lim 
lim 
( x  8)   5
=
= xlim
x3
x3
 3
x3
x3
x  3  x  3  g ( x )  x 4  3x 2  4 x  5 . Thus,
g ( x )  91
x 4  3x 2  4 x  5  91
g  ( 3 )  lim 
lim
=
x3
x  3 = x  3
x3
0
x 4  3x 2  4 x  96 81  27  12  96

 I.F.
lim 
=
x3
33
0
x3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
4
2
By the Factor Theorem, x  3 is a factor of x  3x  4 x  96 . The
4
2
factorization of x  3x  4 x  96 can be found using synthetic division.
4
2
4
3
2
Since x  3x  4 x  96 = x  0 x  3x  4 x  96 , then we write the
following in order to carry out the synthetic division
1
0
3
3
9
4
36
1
3
12
32
 96
96
3
0
4
2
3
2
Thus, x  3x  4 x  96 = ( x  3 ) ( x  3x  12 x  32 ) . Thus,
evaluating the given limit, we have that
3
2
x 4  3x 2  4 x  96
(
x

3
)
(
x

3
x
 12 x  32 )
g  ( 3 )  lim 
lim
=
x3
x3
x3
x3

( x 3  3x 2  12 x  32 )  27  27  36  32  122
= xlim

3
Since the one-sided limits are not equal, then g ( 3 )  xlim
3
g( x )  g( 3)
=
x3
g ( x )  91
x  3.
x3
x  3 = DNE. Thus, the function g is not differentiable at
Thus, g ( 3 ) is undefined.
lim
 6 x 2  36 x , x   2
 3

g
(
x
)

 4x  6x  4 ,  2  x  3
Answer:
 2 x  11
, x3

The graph of the function y  g ( x ) is given below.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
y  g( x ) 
( 3 , g ( 3 ) ) = ( 3 , 91)
(  2 , g (  2 ) ) = (  2 , 31)
3.
h( x )  x 2  36
Using the definition of absolute value given Lesson 2, we have that
 x 2  36 , x 2  36  0
x  36  
. Now, we need to find when
 36  x 2 , x 2  36  0
x 2  36  0 and when x 2  36  0 . We can do this by finding the sign of
x 2  36 using the three-step method from Lesson 1.
2
2
Sign of x  36 :

+
+

6

6
2
2
Thus, x  36  0 when x   6 or x  6 . Since x  36  0 when x   6 ,
2
2
then x  36  0 when x   6 or x  6 .
Also, x  36  0 when
 6  x  6.
 x 2  36 , x   6 or x  6
Thus, h( x )  x  36  
 36  x 2 ,  6  x  6
2
Copyrighted by James D. Anderson, The University of Toledo
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 2 x , x   6 or x  6

h
(
x
)


Thus,
  2x ,  6  x  6
NOTE: Since  6 and 6 are the breakup points of the piecewise function,
then in order to calculate the derivative of the function at x   6 and x  6 ,
we must use the definition of derivative.
To find h(  6 ) , we must calculate h(  6 )  x lim
6
h( x )  h(  6 )
.
x6
h( x )

h
(

6
)

lim
h
(

6
)

0

0
Since
, then we have that
x   6 x  6 . Since
x 2  36
h
(
x
)
lim
h( x )  x 2  36 , then h(  6 )  x lim
6 x  6 = x6
x  6 . Does
this limit look familiar? We discussed this type of limit in Lesson 4. Now, in
x 2  36
order to find x lim
6
x  6 we will have to calculate the one-sided limits
h (  6 ) 
lim
x   6
x 2  36
x6
and h (  6 ) 
lim
x 2  36
x   6
x6 .
Calculating these one-sided limits:
x   6   x   6  h( x )  x 2  36  36  x 2 . Thus,
h (  6 ) 
lim
x   6
lim
x   6
x 2  36
x6
=
lim
x   6
36  x 2
=
x6
lim
x   6
(6  x)(6  x)
=
x6
( 6  x )  12 .
x   6   x   6  h( x )  x 2  36  x 2  36 . Thus,
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
h (  6 ) 
lim
x   6
x 2  36
lim
x   6
x6
=
lim
x   6
x 2  36
=
x6
lim
x   6
( x  6)( x  6)
=
x6
( x  6 )   12 .
Since the one-sided limits are not equal, then h(  6 )  x lim
6
=
lim
h( x )  h(  6 )
x6
x 2  36
x  6 = DNE. Thus, the function h is not differentiable at
x   6 . Thus, h(  6 ) is undefined.
x6
To find h( 6 ) , we must calculate h( 6 )  xlim
6
h( x )  h( 6 )
.
x6
h( x )

h
(
6
)

lim
h
(
6
)

0

0
Since
, then we have that
x6 x  6.
Since
x 2  36
h( x )
lim
h( x )  x  36 , then h( 6 )  xlim
6 x  6 = x6
x  6 . Does this
limit look familiar too? We discussed this type of limit in Lesson 4. Now, in
x 2  36
order to find xlim
6
x  6 we will have to calculate the one-sided limits
2
h ( 6 )  lim 
x6
x 2  36
x6
and h ( 6 )  x lim
6
x 2  36
x6
.
Calculating these one-sided limits:
x  6   x  6  h( x )  x 2  36  x 2  36 . Thus,
h ( 6 )  lim 
x6
x 2  36
x6
( x  6)( x  6)
x 2  36
lim
lim
= x 6
=
x6
x  6 = x  6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
lim ( x  6 )  12 .
x 6
x  6   x  6  h( x )  x 2  36  36  x 2 . Thus,
h ( 6 )  lim 
x6
x 2  36
x6
(6  x)(6  x)
36  x 2
lim
= x lim
=
=
x  6
6
x6
x6
lim [  ( x  6 ) ]   12 .
x 6
Since the one-sided limits are not equal, then h ( 6 )  xlim
6
x 2  36
h( x )  h( 6 )
=
x6
x  6.
x  6 = DNE. Thus, the function h is not differentiable at
Thus, h( 6 ) is undefined.
lim
x6
 2 x , x   6 or x  6

h
(
x
)


Answer:
  2x ,  6  x  6
2
Graph of h( x )  x  36 :
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
Theorem If the function f is differentiable at x  a , then the function f is
continuous at x  a .
f ( x )  f ( a ) . Since the function
Proof We want to show that xlim
a
differentiable at x  a , then f ( a )  xlim
a
f is
f ( x)  f (a)
is a defined real
xa
 f ( x)  f (a) 
f
(
x
)

f ( x) =

 ( x  a )  f ( a ) , then xlim
number. Since
a
x

a


  f ( x)  f (a) 

lim  
(
x

a
)

f
(
a
)
 =

xa
x

a




 f ( x)  f (a) 
lim 
f (a) =
 ( x  a ) + xlim
xa
a
x

a



f ( x)  f (a) 
lim
( x  a )  f ( a ) = f ( a )  0  f ( a )  f ( a ) .
xa
 xlim
a
x

a


Thus, the function f is continuous at x  a .
Continuity does not imply differentiability as our second and third examples of the
piecewise functions above showed. Continuity is necessary but not sufficient for
differentiability for the contrapositive of the theorem above says that if the function
f is discontinuous (not continuous) at x  a , then the function f is not
differentiable at x  a . We could have used this information for our first example
above.
 x 3  64 , x  4
The function f ( x )   3
is discontinuous at x  4 since
 x  27 , x  4
lim f ( x ) = DNE since lim  f ( x ) = lim  ( x 3  27 )  37 and lim  f ( x ) =
x4
x4
x4
x4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850
lim  ( x 3  64 )  128 . Of course, we can see from the graph of y  f ( x )
x4
below, that the function f is discontinuous at x  4 :
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1850