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3B MAS
4. Functions
Limit of a Function
Graphically the limiting value of a function
f(x) as x gets closer and closer to a certain
value (say 'a') is obtained by moving along
the curve from both sides of 'a' as x moves
toward 'a'.
The limiting value of f(x) as x gets closer
and closer to 'a' is denoted by
lim f (x)
x a
Right/Left Hand Limits
As x moves towards 'a' from right (left)
hand side, the limiting value of f(x) is
denoted by
lim f ( x ) and
x a
lim  f ( x )
x a
Limiting and Functional Value
If both sides limits are equal,
lim f ( x)  lim f ( x )  lim f ( x )
x a
x a
x a
Otherwise, lim f (x) does not exist.
xa
Note that lim f (x) may not equal to f(a)
xa
Limiting Value
y
f(x)
lim f ( x)  f (a)
x a
f(a)
a
x
Example 1
Find the limit of the f(x) as x approaches a for the
following functions.
(a)
The limit does not
exist as the function
is not defined 'near' a.
a
Example 1 (cont'd)
(b)
The limit does not exist
as the left side limit is
not the same as right
side limit.
a
Example 1 (cont'd)
(c)
The limit exists but it
does not equal to f(a).
a
f(a)
Example 1 (cont'd)
(d)
The limit exists and it
equals to f(a).
lim f (x)  f (a)
f(a)
x a
a
Evaluating Limits
If f(x) is not broken at 'a', use direct
substitution to evaluate its limit as x
approaches 'a'
Otherwise, find the left side and right side
limits and check if they are equal.
Example 2
Evaluate the following limits if they exist.
(a) f(x) = 2x – 5 as x  1
f(x) is not broken at x = 1, so use direct sub.
lim (2x  5)  2  5  3
x 1
Example 2 (cont'd)
(b) f(x) = ln x as x  0
f(0) is not defined. So consider limit from both sides.
lim  ln x  
x 0
But f(x) is not defined for x < 0.
So the limit does not exist.
Example 2
(c) f(x) = 1/(x – 2) as x  2
f(2) is not defined. So consider limit from both sides.
lim 1 /( x  2)   and
x 2
lim  1 /( x  2)  
x 2
Since the left side limit does not equal to the right
side limit, the limit of the f(x) as x approaches 2
does not exist.
Example 2 (cont'd)
(d) f(x) = (x – 1)/(x2 – 1) as x  1
f(x) = (x – 1)/(x + 1)(x – 1) = 1/(x + 1)
1/(x + 1) is not broken at x = 1, so use direct sub.
x 1
lim 2
 0.5
x 1 x  1
Example 2 (cont'd)
 x 1
(e) f (x)   2
x  7
if x  3
if x  3
lim f (x)  lim (x  1)  2
x 3
x 3
lim f (x)  lim (x 2  7)  2
x 3
x 3
So lim f (x)  2
x 3
Ex 6C: odd numbers
Limits to Infinity
If f(x) = x + c, f(x)   as x   (note
that x is the dominant term)
If f(x) = 1/x, f(x)  0 as x  
If f(x) = ax2 + bx + c, ax2 is the dominant
term as x  
b c
ax  bx  c  x (a   2 )
x x
2
2
Example 3
Find the limit of f(x) as x   (if they exist) for:
(a )
( b)
(c)
(d )
2x  3
f (x) 
4x  1
1  7x
f (x)  2
x  2x  3
x 2  2x  4
f (x) 
 x2  x 1
3x 3  7 x 2  2
f (x)  2
x  5x  4
Example 3 (cont'd)
2x  3 2x 1
(a ) f ( x ) 


4x  1 4x 2
2x  3 1
So lim

x  4 x  1
2
as x  
1  7x
 7x
7
( b) f ( x )  2
 2 
x
x  2x  3
x
1  7x
So lim 2
0
x  x  2 x  3
as x  
Example 3 (cont'd)
x 2  2x  4
x2
(c) f ( x ) 

 1 as x  
2
2
 x  x 1  x
x 2  2x  4
So lim
 1
2
x    x  x  1
3 x 3  7 x 2  2 3x 3
(d ) f ( x )  2
 2  3x as x  
x  5x  4
x
So lim f ( x )   and lim f ( x )  
x  
Ex 6D: even numbers
x 
Trigonometric Limits
y
lim cos x  1
2
x 0
lim sin x  0
y = tan x
y = cos x
x 0
lim tan x  0
x 0
y = sin x
1.5
1
0.5
-180
-90
90
-0.5
-1
-1.5
-2
180
270
360
x
Example 4
Find the following limits.
sin x
(a ) lim
x 0 x
tan x
(b) lim
x 0
x
1  cos x
(c) lim
x 0
x
tan 4 x
(d ) lim
x 0
x
Example 4 (cont'd)
Area of OAC = r2 sin x / 2
Area of sector OAC = r2 x / 2
Area of OAB = r2 tan x / 2
B
C
So (size of areas)
sin x < x < tan x
1 < x / sin x < 1 / cos x
1 > sin x / x > cos x
Take limit as x  0 to get
O
sin x
lim
1
x 0 x
That means sin x  x as x  0
x
r
A
Example 4 (cont'd)
tan x
sin x 1
(b) lim
 lim
1
x 0 x
x 0 x cos x
1  cos x
(c) lim
x 0
x
1  cos x 1  cos x
1  cos 2 x
 lim
 lim
x 0
x
1  cos x x 0 x (1  cos x )
sin x sin x
 lim
 1 0
x 0 x 1  cos x
0
tan 4 x
tan 4 x
tan 4 x
(d ) lim
 4 lim
 4 lim
4
x 0
x 0 4 x
4 x 0 4 x
x
Continuity
Graphically a graph is continuous at x = a if
it is not broken (disconnected) at that point.
Algebraically the limit of the function from
both sides of 'a' must equal to f(a).
lim f ( x )  lim f ( x )  f (a )
x a
x a 
Example 5
The following functions are not continuous at x = a.
Why?
a
a
f(a)
a
Example 6
The following functions are continuous at x = a.
f(a)
a
a
Example 7
Determine if the given function is continuous at the
given point.
(a) f(x) = | x – 2 |
at x = 2
(b) f(x) = x
at x = 0
(c) f(x) = 1 / (x + 3)
at x = -3
Example 7 (cont'd)
(a )
lim | x  2 |  lim | x  2 |  0  f (2)
x 2
x 2
So | x  2 | is continuous at x  2.
( b)
x is not defined for x  0
So
x
is not continuous at x  0.
1
1
(c) lim 
  but lim 
 
x  3 x  3
x  3 x  3
So 1 /( x  3) is not continuous at x  3.
Example 8
Given that f(x) is continuous over the set of all real
numbers, find the values of a and b.
x 2  a
x  1

f ( x )   bx  6  1  x  2
2 x  a
x2

Example 8 (cont'd)
Only need to consider the junctions (x = -1 and x = 2)
f (1)  1  a  lim  f ( x ) and
x 1
lim  f ( x )  b  6
x 1
So 1  a  b  6
f (2)  2b  6  lim  f ( x ) and
x 2
So  2b  6  4  a
 a  4 and
b  1
Ex 6E: odd numbers
lim  f ( x )  4  a
x 2
Differentiability
Graphical approach: A function f(x) is said to
be differentiable at x = a if there is no 'corner'
or 'vertical tangency' at that point.
A function must be continuous (but not
sufficient) in order that it may be
differentiable at that point.
Example 9
The following functions are not differentiable at x = a.
(a)
Corner at x = a
f(a)
a
Example 9 (cont'd)
y
2
(b)
1
– 4
– 3
– 2
– 1
1
– 1
– 2
Vertical tangency at x = 1
2
3
4
5
x
Example 9 (cont'd)
y
8
(c)
6
4
2
–5
–4
–3
–2
–1
–2
–4
–6
–8
Not continuous (not even defined) at x = -2
1
x
Example 10
The following functions are differentiable everywhere.
(a)
Example 10 (cont'd)
(b)
Example 10 (cont'd)
(c)
Derivative of a Function
A function is differentiable at a point if it is
continuous (not broken), smooth (no corner)
and not vertical (no vertical tangency) at
that point.
Its derivative is given by (First Principle)
f ( x  h)  f ( x )
lim
h 0
h
Differentiability (cont'd)
Q
f(x+h)
f(x+h) – f(x)
f(x)
h
x
x+h
Differentiability (cont'd)
The gradient of PQ is given by
f ( x  h)  f ( x )
h
As Q moves closer and closer to P (i.e. as h
tends to 0), the limiting value of the
gradient of PQ (i.e. the derivative of f(x) at
x) becomes the tangent at P.
Differentiability (cont'd)
The derivative of a function y = f(x) is
denoted by
dy
df
or y or
or f 
dx
dx
It also represents the rate of change of y
with respect to x.
Example 11
(a) Find the gradient function of y = 2x2 using first
principle. Find also the gradient at the point
(3, 18).
(b) Use the definition (first principle) to find the
derivative of ln x and hence find the derivative of
ex.
Example 11 (cont'd)
dy
2( x  h ) 2  2 x 2
(a )
 lim
dx h 0
h
2( x 2  2hx  h 2 )  2 x 2
 lim
h 0
h
4hx  2h 2
 lim
 lim (4 x  2h )
h 0
h 0
h
 4x
dy
dx
 4  3  12
x 3
Example 11 (cont'd)
xh
ln
d
ln (x  h)  ln x
x
ln x  lim
 lim
h 0
h 0
dx
h
h
1
h
1
x
h
 lim ln (1  )  ln [lim (1  )]
h 0 h
h 0 h
x
x
x
1
1 m
x
 ln [ lim (1  ) ]
(where m  )
m 
x
m
h
1
 ln e
x
1

x
Example 11 (cont'd)
ye
x
 ln y  x
d
d
d
ln y  1 
ln y  y  1
dx
dy
dx
1 dy
dy

1 
y
y dx
dx
d x
x

e e
dx
Example 12
Find the derivative of the following functions from
first principles.
(a) f(x) = 1/x
(b) f(x) = x
(c) f(x) = xn
Example 12 (cont'd)
(a )
1
1

d 1
( )  lim x  h x
h 0
dx x
h
xxh
x(x  h)
 lim
h 0
h
1
 lim
h 0 x ( x  h )
1
 2
x
Example 12 (cont'd)
( b)
d
dx
xh  x
x  lim
h 0
h
xh  x
xh  x
 lim

h 0
h
xh  x
xhx
 lim
h 0 h ( x  h 
x)
1
 lim
h 0
xh  x
1

2 x
Example 12 (cont'd)
d n
(x  h)  x
x  lim
h 0
dx
h
nx n 1h  n C 2 x n  2 h 2  ......  h n
 lim
h 0
h
n 1
n 2
n 1
 lim (nx  n C 2 x h  ......  h )
n
(c)
h 0
 nx n 1
n
Concavity
If f(x) opens downward, it is said to be
concave down
If f(x) opens upwards, it is concave up
concave down
f '(x): +
0
-
concave up
-
0
+
Point of Inflection
Points of inflection: points where the curve
changes from concave up to concave down
or concave down to concave up
point of inflection
f '(x):
- - -
+++
maximum
Horizontal Inflection
Horizontal inflection: a point of inflection
where the graph is momentarily horizontal,
dy/dx = 0
+ve
-ve
+ve
-ve
horizontal inflection
Stationary Points
Turning points: max and min
Stationary points: max, min and horizontal
inflection
dy/dx = 0
Example 13
Given the f(x) graph below draw f '(x).
y
x
Example 13 (cont'd)
y
f '(x) changes as below:
x < a: +ve, but 
x = a: f '(a) = 0
local max
a
f (x)
b
x
c
f '(x)
a < x < b:
-ve, , then (less –ve)
point of inflection
Example 13 (cont'd)
y
x = b: f '(b) = 0
local min
b < x < c:
+ve, , then 
point of inflection
a
f (x)
b
x
c
f '(x)
x = c: f '(c) = 0
global max
x > c: -ve, 
Noteworthy Features
Min TP: dy/dx = 0, sign change –ve, 0, +ve
Max TP: dy/dx = 0, sign change +ve, 0, –ve
Horizontal inflection: dy/dx = 0, +ve, 0, +ve
or –ve, 0, –ve, (i.e. no sign change)
Point of inflection: d2y/dx2 = 0, (dy/dx is a
max/min),
Example 14
Graph the following function and its derivative.
Use your graphs to locate the stationary points and
points of inflection on
y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5
and determine the nature of each.
Example 14 (cont'd)
y = x4/4 – 4x3/3 – 7x2/2 + 10x + 5
dy/dx = x3 – 4x2 – 7x + 10
TP  dy/dx = 0
So x3 – 4x2 – 7x + 10 = 0
i.e. (x + 2)(x – 1)(x – 5) = 0
 x = -2, 1 or 5
When x = -2, y = -43/3
When x = 1, y = 125/12
When x = 5, y = -515/12
Example 14 (cont'd)
y
20
10
-4
-2
max
PoI
f(x)
2
4
6
-10
min
PoI
-20
f '(x)
-30
-40
min
8
x
Piecewise Defined Functions
A piecewise defined function has different
formulas for different parts of its domain.
At junction a filled circle indicates that a
point actually exists there, whereas an
empty circle shows a discontinuous point.
Example 15
Given the function below
1 / x  1 for x  0
 2
f ( x )  x
for 0  x  2
x  1
for x  2

(a) Find f(-2), f(1) and f(2)
(b) Graph f and determine whether f is continuous
at x = 0 and x = 2.
Example 15 (cont'd)
1 / x  1 for x  0
 2
f ( x )  x
for 0  x  2
x  1
for x  2

(a) f(-2) = 1/(-2) – 1 = -3/2
f(1) = 12 = 1
f(2) = 2 + 1 = 3
Example 15 (cont'd)
y
 lim f ( x )  lim f ( x )
x 0
6
x 0
5
f ( x ) is not continuous at x  0
4
 lim f ( x )  lim f ( x )
x 2
3
x 2
2
f ( x ) is not continuous at x  2
–4
–3
–2
1
–1
–1
–2
–3
–4
1
2
3
4 x
Example 16
Graph y = | x2 – 6x + 8 | and determine whether the
function is continuous at x = 2 and x = 4.
Example 16 (cont'd)
y
8
From graph, the
function is
continuous at
x = 2 and x = 4.
6
4
2
-1
1
-2
2
3
4
5
6
7
x
The Sign Function
It can be considered as a logical function
(especially in computer science)
It extracts the sign of the function
It returns 1 if f(x) is positive, 0 if f(x) equals
to 0 and –1 if f(x) is negative.
1

sgn ( f (x) )  0
1

f (x)  0
f (x)  0
f (x)  0
Sign Function (cont'd)
y
2
1

sgn (f (x))  0
1

1
-3
-2
-1
1
-1
-2
2
3
x
f (x)  0
f (x)  0
f (x)  0
Example 17
Graph
(a) y = sgn (x/|x|)
(b) y = sgn (x2 – 1)
Example 17 (cont'd)
(a) y = sgn (x/|x|)
y
2
1
–3
–2
–1
1
–1
–2
2
3
Not continuous at
x=0
x
Example 17 (cont'd)
(b) y = sgn (x2 – 1)
y
2
1
–2
–1
Not continuous at
x = -1 and x = 1
1
–1
–2
2
x
Greatest Integer Function
Also known as floor function
Defined as the greatest integer less than or
equal to the number
That is, it rounds any number down to the
nearest integer
Symbol: int [x] or x 
int [4.2] = 4.2 = 4
int [-2.1] =  2.1 = -3
Greatest Integer Function (cont'd)
y
3
2
1
–3
–2
–1
1
2
3
4
x
–1
–2
–3
Not continuous at
all integers.
Example 18
Graph the following functions:
(a) int [2x – 1]
(b) int [x2]
Example 18 (cont'd)
y
(a) int [2x –1]
3
2
Consider 2x – 1 = n
where n is an integer
x = (n + 1) / 2
So the 'breaking
points' are steps of
half of an integer
1
–2
–
3
2
–1
–
1
1
2
–1
2
–2
–3
–4
1
3
2
2
5
2
x
Example 18 (cont'd)
y
4
(b) int [x2]
3
Consider x2 = n
where n is a positive
integer
x = n
So the breaking
points are  square
root of +ve integers
2
1
–
5 –2
2
–
3 –1
2
–
1
1
2
2
1
3
2
2
5
2
x
Differentiation
dy
f (x  h)  f (x)
 lim
dx h 0
h
Simple Power Function
If
y  kx n
dy
n 1
 knx
dx
where k and n are real numbers
Example 19
Find the derivative of the following functions:
(a ) y  4 x
( b) y  x
2
(c) y  6
(d ) P  8 t 3
(e) C  q
1
( f ) y  3x  4
(g ) y  5 x
Example 19 (cont'd)
(a)
y  4x
(b)
yx
(c)
y6
(d)
P  8t 3
2
dy
 4  2x 21  8x
dx
dy
 11x11  1
dx
dy
 6  0x 01  0
dx
dP
 8  3t 31  24t 2
dt
Example 19 (cont'd)
1
(e)
Cq
(f )
y  3x 4
(g)
y5 x
dC
1
11
 1 ( 1)q
 2
dq
q
dy
12
4 1
 3  ( 4)x
 5
dx
x
dy
1 12 1
5
 5 x

dx
2
2 x
Sum/Difference of Functions
If
y  f ( x )  g( x )
dy d
d
 f ( x )  g( x )
dx dx
dx
i.e. y'  f ' ( x )  g ' ( x )
Example 20
Find the derivative of the following functions:
3
(a ) y  7 x 
x
(b) y  3x 2  x  10
Example 20 (cont'd)
3
(a ) y  7 x 
x
y'  7  3(1) x
2
3
7 2
x
(b) y  3x 2  x  10
y'  3(2) x  1  0  6x  1
The Chain Rule
If
y  f (u ) and u  g( x )
dy df du
 
dx du dx
[i.e. If
y  f (g ( x ) )
y'  f ' (g( x ) )  g' ( x )]
Example 21
Find the derivative of the following functions:
(a ) y  u 3 and u  2x  3 (or y  (2x  3)3 )
4
4
2
(b) y  2 and u  x  3x  1 (or y  2
)
2
u
( x  3x  1)
Example 21 (cont'd)
(a ) y  u 3 and u  2x  3 (or y  (2x  3)3 )
dy
du
2
 3u and
2
du
dx
dy
2

 3(2x  3)  2
dx
2
 6(2x  3)
Example 21 (cont'd)
4
4
2
(b) y  2 and u  x  3x  1 (or y  2
)
2
u
( x  3x  1)
dy
du
3
 4(2)u and
 2x  3
du
dx
dy

 8( x 2  3x  1) 3  (2x  3)
dx
 8(2x  3)
 2
( x  3x  1)3
The Product Rule
If
y  f ( x ) g( x )
then
dy
 f ' ( x ) g( x )  f ( x ) g' ( x )
dx
Example 22
Differentiate the following functions:
(a )
y  (3x  4) 2 (7 x  2)
( b)
5
4
y
(2 x  3)
x 8
Example 22 (cont'd)
(a )
y  (3x  4) 2 (7 x  2)
y'  2(3x  4)(3)(7 x  2)  (3x  4) 2 (7)
 6(21x 2  22 x  8)  7(9 x 2  24x  16)
 189 x 2  300 x  64
Example 22 (cont'd)
( b)
5
y
(2 x  3) 4  5( x  8) 1 (2 x  3) 4
x 8
y'  5(1)( x  8)  2 (2 x  3) 4  5( x  8) 1 4(2 x  3) 3 (2)
 5(2 x  3) 4 40(2 x  3) 3


2
( x  8)
x 8
5(2 x  3) 3
2x  3

[8 
]
x 8
x 8
5(2 x  3) 3 (6 x  67)

( x  8) 2
The Quotient Rule
If
f (x)
y
g(x )
then
dy f ' g  f g '

dx
g2
Example 23
Differentiate the following functions:
(a )
x 2  4x  7
y
3x  2
( b)
(9 x  4) 2
y
3x 3  x  1
Example 23 (cont'd)
(a )
x 2  4x  7
y
3x  2
dy ( 2 x  4)(3x  2)  ( x 2  4 x  7)(3)

dx
(3x  2) 2

dy 3x 2  4 x  29

dx
(3x  2) 2
Example 23 (cont'd)
( b)
(9 x  4) 2
y
3x 3  x  1
dy 2(9 x  4)(9)(3x 3  x  1)  (9 x  4) 2 (9 x 2  1)

dx
(3x 3  x  1) 2

dy 18(9 x  4)(3x 3  x  1)  (9 x  4) 2 (9 x 2  1)

dx
(3x 3  x  1) 2
Second Derivatives
If
y  f (x)
then
dy
 f ' ( x ) is called the first derivative
dx
and
d  dy  d 2 y
   2  f ' ' (x)
dx  dx  dx
is called the 2nd derivative
Example 24
Find the second derivative of the following functions
(a ) y  (6 x  7) 3
( b) y  4 x  2 x  9
2
Example 24 (cont'd)
(a ) y  (6 x  7 ) 3
dy
 18(6x  7) 2
dx
and
d2y
 36(6x  7)(6)
2
dx
( b) y  4 x 2  2 x  9
dy
 8x  2 and
dx
2
d y
8
2
dx
Example 25
Find the derivative of the following functions:
3
(a ) y  7 x 
x
4
( b) y  2
( x  3x  1) 2
5
(c) y 
( 2 x  3) 4
x 8
(9 x  4) 2
(d ) y  3
3x  x  1
Example 25 (cont'd)
3
(a ) y  7 x 
x
3
y'  7  3(1) x  7  2
x
(b) y  4( x 2  3x  1) 2
2
y '  4  (2)  ( x 2  3x  1) 3 (2x  3)
 8(2x  3)
 2
( x  3x  1)3
Example 25 (cont'd)
(c)
5
y
(2 x  3) 4  5( x  8) 1 (2 x  3) 4
x 8
y'  5(1)( x  8) 2 (2 x  3) 4  5( x  8) 1 4(2 x  3)3 (2)
 5(2 x  3) 4 40(2 x  3)3


2
x 8
( x  8)
5(2 x  3)3
2x  3

[8 
]
x 8
x 8
5(2 x  3)3 (6 x  67)

( x  8) 2
Example 25 (cont'd)
(d )
(9 x  4) 2
y 3
3x  x  1
dy 2(9 x  4)(9)(3x 3  x  1)  (9x  4) 2 (9x 2  1)

3
2
dx
(3x  x  1)

dy 18(9x  4)(3x 3  x  1)  (9x  4) 2 (9x 2  1)

dx
(3x 3  x  1) 2
f(x)
Differentiating e
and ln f(x)
d f (x)
e
 f ' (x)  ef ( x )
dx
d
f ' (x)
ln f ( x ) 
dx
f (x)
Example 26
Differentiate the following with respect to x:
(a )
( b)
y  3 ex  e
y2e
2 x 1
1
 3x
2e
Example 26 (cont'd)
(a ) y  3 e x  e
dy
x
3e
dx
1
1 3 x
2 x 1
( b) y  2 e
 3x  2 e
 e
2e
2
dy
 3 3 x
2 x 1
 2(2) e
( )e
dx
2
dy
3 3 x
2 x 1

4e
 e
dx
2
2 x 1
Example 27
Differentiate
1
(a ) y  ln ( x 2  7 x  3)
e
2
x
e
x
(b) y  e ln
1  ex
Example 27 (cont'd)
1
(a ) y  ln ( x 2  7 x  3)
e
dy 1 2 x  7
 ( 2
)
dx e x  7 x  3
2
x
e
2x
2x
2
x
(b) y  e ln

e
[ln
(
x

e
)

ln
(
1

e
)]
x
1 e
x
2
x
e
2x
2
x
2x
 2e [ln ( x  e)  ln (1  e )]  e [ 2

]
x
x  e 1 e
2
x
dy
x

e
2
x
e
 e 2 x [2 ln (
) 2

]
x
x
dx
1 e
x  e 1 e
Differentiability (Revisit)
Graphical approach: continuous, no corner,
no vertical tangency
Algebraical approach:
lim f ( x )  lim f ( x )  f (a ) (continuous )
x a
x a
lim f ( x )  lim f ( x )  f (a ) (no corner )
'
x a
'
f (a ) is finite
'
'
x a
(no vertical tan gency )
Example 28
Determine if the following functions are differentiable
at the indicated points.
(a) y = 1 / (x + 1)
at x = -1
(b) y = | x + 1 |
at x = -1
(c) f(x) = -6x + 5 for x < 3
= -x2 – 4 for x  3
at x = 3
Example 28 (cont'd)
(a) Let f(x) = 1 / (x + 1)
f(-1) is not defined
f(x) is not continuous at x = -1
f(x) is not differentiable at x = -1
Example 28 (cont'd)
(b) Let f ( x )  | x  1 |
 x  1 x  1
f (x)  
x 1
x  1
lim  f ( x )  lim  f ( x )  f (1)  0
x  1
x  1
 f ( x ) is continuous at x  1
 1
f ' (x)  
1
x  1
x 1
lim  f ' ( x )  1  lim  f ' ( x )  1
x  1
x  1
 f ( x ) is not differenti able at x  1
Example 28 (cont'd)
 6 x  5 x  3
(c) f ( x )   2
 x  4 x  3
lim f ( x )  lim f ( x )  f (3)  13
x 3
x 3
 f ( x ) is continuous at x  3
x3
 6
f ' (x)  
x 3
 2 x
lim f ' ( x )  lim f ' ( x )  6
x 3
x 3
 f ( x ) is differenti able at x  3
Example 29
Find the value of a and b so that
f(x) = 3x + 1
for x < 1
= x2 + ax + b for x  1
is continuous and differentiable everywhere.
Example 29 (cont'd)
Possible discontinuity and non-differentiability at x = 1
continuous at x = 1 if
i.e. a + b = 3
4=1+a+b
f ’(x) = 3 for x < 1
= 2x + a
for x > 1
differentiable at x = 1 if 3 = 2 + a
So
a = 1 and b = 2
Riemann Sums
To find an approximate area under a curve
between two x values [a, b]
The area is divided into n rectangles of
equal width
So the width x = (b – a)/n
There are many ways to find the height h of
each rectangle (see later)
Then the required area A =  hx over the
interval [a, b]
Example 30
The shaded area below shows the exact area under the
curve f(x) = x3 – 3x2 + 8 in the interval [0, 3]
y
9
8
7
6
y = x3 – 3x2 + 8
5
4
3
Actual area = 17.25
2
1
0.5
1
1.5
2
2.5
3
x
Example 31
(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6
and h = left endpoint
Left endpoint for the
3rd rectangle
x = (3 – 0)/5 = 0.6
Example 31 (cont'd)
f(x1)= (0)3 – 3(0)2 + 8 = 8
f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136
f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408
f(x4)= (1.8)3 – 3(1.8)2 + 8 = 4.112
f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544
5
A   f ( x i ) x
i 1
 0.6(8  7.136  5.408  4.112  4.544)
 17.52
Example 32
(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6
and h = right endpoint
Right endpoint for the
3rd rectangle
Example 32 (cont'd)
f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136
f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408
f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112
f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544
f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8
5
A   f ( x i ) x
i 1
 0.6(7.136  5.408  4.112  4.544  8)
 17.52
Example 33
(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6
and h = minimum point
Minimum point for the 4th
rectangle
Example 33 (cont'd)
f(x1)= (0.6)3 – 3(0.6)2 + 8 = 7.136
f(x2)= (1.2)3 – 3(1.2)2 + 8 = 5.408
f(x3)= (1.8)3 – 3(1.8)2 + 8 = 4.112
f(x4)= (2.0)3 – 3(2.0)2 + 8 = 4
f(x5)= (2.4)3 – 3(2.4)2 + 8 = 4.544
5
A   f ( x i ) x
i 1
 0.6(7.136  5.408  4.112  4  4.544)
 15.12
Example 34
(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6
and h = maximum point
Maximum point for the 4th
rectangle
Example 34 (cont'd)
f(x1)= (0)3 – 3(0)2 + 8 = 8
f(x2)= (0.6)3 – 3(0.6)2 + 8 = 7.136
f(x3)= (1.2)3 – 3(1.2)2 + 8 = 5.408
f(x4)= (2.4)3 – 3(2.4)2 + 8 = 4.544
f(x5)= (3.0)3 – 3(3.0)2 + 8 = 8
5
A   f ( x i ) x
i 1
 0.6(8  7.136  5.408  4.544  8)
 19.8528
Example 35
(Example 24) Consider n = 5, x = (3 – 0)/5 = 0.6
and h = midpoint
Midpoint for the 3rd
rectangle
Example 35 (cont'd)
f(x1)= (0.3)3 – 3(0.3)2 + 8 = 7.757
f(x2)= (0.9)3 – 3(0.9)2 + 8 = 6.299
f(x3)= (1.5)3 – 3(1.5)2 + 8 = 4.625
f(x4)= (2.1)3 – 3(2.1)2 + 8 = 4.031
f(x5)= (2.7)3 – 3(2.7)2 + 8 = 5.813
5
A   f ( x i ) x
i 1
 0.6(7.757  6.299  4.625  4.031  5.813)
 17.115
A Better Approximation
Due to the use of h (left, right, mid, min and
max), the rectangles do not truly represent
the area under the curve for each strip
If n (number of rectangles) increases, the
error decreases
Example 36
n = 20, x = (3 – 0)/20 = 0.15, h = midpoint
Actual area = 17.25
Example 37
n = 100, x = (3 – 0)/100 = 0.03, h = midpoint
Actual area = 17.25
Limit of a Sum
The more rectangles, the greater accuracy
So the actual area A is given by
A  lim
n 
 f (x)x
This is written as
A  lim
n 
where

 f (x)x   f (x)dx
represents the lim it of a sum
and the above process is called int egration
Integration
 f(x)dx represents the area under the curve
y
8
f(x)
7
4
 f (x)dx
6
1
5
4
3
2
1
1
2
3
4
x
Example 38
Evaluate the area under the curve y = 3 from x = 2 to
x = b by completing the following table.
b
3
4
5
6
Area
Hence give an answer for
where k is a constant

x
a
kdx
7
x
Example 38 (cont'd)
b
3
4
5
6
7
b
Area 3
6
9
12 15 3(b – 2)
y
4
3

b
a
kdx  k(b  a )
2
1
1
2
3
4
5
6
7
8
9
10
x
Example 39
Evaluate the area under the curve y = 2x from
x = 0 to x = b by completing the following table.
b
0
1
2
3
Area
Hence give an answer for
where k is a constant

b
0
kxdx
4
b
Example 39 (cont'd)
b
0
1
2
3
4
x
Area 0
1
4
9
16 b2b/2 = b2
y
14
12

b
0
kb
kxdx 
2
2
10
8
6
4
2
1
2
3
4
5
6
x
Example 40
Given the areas under the curve y = x2 from
x = 0 to x = 4 in the following table, find the area
when x = b
b
0
Area
0
1
2
1/3 8/3
Hence give an answer for

b
0
3
4
9
64/3
x 2dx
b
Example 40 (cont'd)
b
0
1
2
3
4
b
Area 0 1/3 8/3 9 64/3 b3/3
y
30
25

b
0
3
b
x dx 
3
20
2
15
10
5
1
2
3
4
5
x