P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
C H A P T E R
19
E
Differentiation of
Polynomials
Objectives
PL
P1: FXS/ABE
To understand the concept of limit.
To understand the definition of differentiation.
To understand and use the notation for the derivative of a polynomial function.
To be able to find the gradient of a curve of a polynomial function by calculating
its derivative.
M
To apply the rules for differentiating polynomials to solving problems.
SA
In the previous chapter the rate of change of one quantity with respect to a second quantity has
been considered. In this chapter a technique will be developed for calculating the rate of
change for polynomial functions. To illustrate this an introductory example will be considered.
On planet X, an object falls a distance of y metres in t seconds, where y = 0.8t2 .
(Note: On Earth the commonly used model is y = 4.9t2 .)
Can a general expression for the speed
of such an object after t seconds be
found? In the previous chapter it was
y
found that the gradient of the curve
y = 0.8t 2
at a given point P can be approximated
by finding the gradient of a chord PQ,
Q
where Q is a point on the curve as close
as possible to P.
The gradient of chord PQ approximates
P
the speed of the object at P. The
‘shorter’ this chord is made the better
t
5
(5 + h)
0
the approximation.
536
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
537
Chapter 19 — Differentiation of Polynomials
Let P be the point on the curve where t = 5. Let Q be the point on the curve corresponding
to h seconds after t = 5, i.e. Q is the point on the curve where t = 5 + h.
0.8 (5 + h)2 − 52
The gradient of chord PQ =
(5 + h) − 5
0.8 (5 + h)2 − 52
=
h
= 0.8(10 + h)
E
The table gives the gradient for different values of h.
Use your calculator to check these.
If values of h of smaller and smaller magnitude are
taken, it is found that the gradient of chord PQ gets
closer and closer to 8. At the point where t = 5 the
gradient is 8. Thus the speed of the object at the
moment t = 5 is 8 metres per second.
The speed of the object at the moment t = 5 is the
limiting value of the gradients of PQ as Q approaches P.
Gradient of PQ
0.7
0.6
0.5
0.4
0.3
0.2
0.1
8.56
8.48
8.40
8.32
8.24
8.16
8.08
PL
h
M
A formula for the speed of the object at any time t is required. Let P be the point with
coordinates (t, 0.8t2 ) on the curve and Q be the point with coordinates (t + h, 0.8(t + h)2 ).
0.8 (t + h)2 − t 2
The gradient of chord PQ =
(t + h) − t
= 0.8(2t + h)
SA
From this an expression for the speed can be found. Consider the limit as h approaches 0, that
is, the value of 0.8(2t + h) as h becomes arbitrarily small.
The speed at time t is 1.6t metres per second. (The gradient of the curve at the point
corresponding to time t is 1.6t.) This technique can be used to investigate the gradient of
similar functions.
19.1
The gradient of a curve at a point, and the
gradient function
Consider the function f : R →R, f (x) = x 2 .
The gradient of the chord PQ in the
adjacent graph
y
f (x) = x2
(a + h, (a + h)2)
(a + h)2 − a 2
=
a+h−a
=
a 2 + 2ah + h 2 − a 2
a+h−a
= 2a + h
Q
P(a, a2)
0
x
and the gradient at P can be seen to be 2a. The limit of (2a + h) as h approaches 0 is 2a.
It can be seen that there is nothing special about a. So if x is a real number a similar formula
holds.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
538
Essential Mathematical Methods 1 & 2 CAS
E
The gradient of the function y = x2 at any point x is equal to 2x.
It is said that 2x is the derivative of x2 with respect to x or, more briefly, the derivative of
2
x is 2x.
This itself is clearly the rule for a function of x and we refer to this function as the gradient
(or derived) function.
The straight line that passes through P and has gradient 2a is called the tangent to the curve
at P.
From the discussion at the beginning of the chapter it can be seen that the derivative of 0.8t2
is 1.6t.
Example 1
By first considering the gradient of the chord QP, find the gradient of y = x2 − 2x at the point
Q with coordinates (3, 3).
Solution
PL
P1: FXS/ABE
y
Consider chord PQ:
Gradient of PQ
y = x2 – 2x
(3 + h, (3 + h)2 – 2(3 + h))
(3 + h)2 − 2(3 + h) − 3
3+h−3
9 + 6h + h 2 − 6 − 2h − 3
=
3+h−3
2
4h + h
=
h
=4+h
M
=
0
P
Q(3, 3)
2
Consider h as it approaches zero. The gradient at the point (3, 3) is 4.
SA
Example 2
Find the gradient of chord PQ and hence the derivative of x2 + x.
y
y = x2 + x
P(x + h, (x + h)2 + (x + h))
–1
Q(x, x2 + x)
x
0
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
x
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
539
Solution
(x + h)2 + (x + h) − (x 2 + x)
x +h−x
2
x + 2xh + h 2 + x + h − x 2 − x
=
h
2xh + h 2 + h
=
h
= 2x + h + 1
E
The gradient of chord PQ =
From this it is seen that the derivative of x2 + x is 2x + 1.
The notation for limit of 2x + h + 1 as h approaches 0 is lim 2x + h + 1.
h→0
The derivative of a function with rule f (x) may be found by first finding an expression for
the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) and then finding the limit of
this expression as h approaches 0.
Example 3
PL
P1: FXS/ABE
By first considering the gradient of the chord from Q(x, f (x)) to P(x + h, f (x + h)) for the
curve f (x) = x 3 , find the derivative of x3 .
M
Solution
f (x) = x 3
f (x + h) = (x + h)3
The gradient of chord PQ =
SA
=
f (x + h) − f (x)
(x + h) − x
(x + h)3 − x 3
(x + h) − x
(x + h)3 − x 3
h→0 (x + h) − x
The derivative of f (x) = lim
x 3 + 3x 2 h + 3h 2 x + h 3 − x 3
h→0
h
2
2
3x h + 3h x + h 3
= lim
h→0
h
= lim
= lim 3x 2 + 3hx + h 2
h→0
= 3x 2
We have found that the derivative of x3 is 3x2 . The following example provides practice in
determining limits.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
540
Essential Mathematical Methods 1 & 2 CAS
Example 4
Find:
a lim 22x 2 + 20xh + h
b lim
c lim 3x
d lim 4
h→0
3x 2 h + 2h 2
h→0
h
h→0
h→0
Solution
a lim 22x 2 + 20xh + h = 22x 2
h→0
c lim 3x = 3x
b lim
d lim 4 = 4
h→0
PL
h→0
3x 2 h + 2h 2
= lim 3x 2 + 2h
h→0
h→0
h
= 3x 2
E
P1: FXS/ABE
Using the TI-Nspire
SA
M
Use Define (b 1 1) with the function
√
f (x) = x and then complete as shown.
Using the Casio ClassPad
For Example 4b, enter and highlight the
expression (3x 2 h + h 2 )/ h then from the
and enter h and
keyboard select 2D—Calc—
0 in the spaces provided, as shown.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
541
Exercise 19A
1 A space vehicle moves so that the distance travelled over its first minute of motion is given
by y = 4t4 , where y is the distance travelled in metres and t the time in seconds. By finding
the gradient of the chord between the points where t = 4 and t = 5, estimate the speed of
the space vehicle when t = 5.
4
3 Find:
2x 2 h 3 + xh 2 + h
h→0
h
b lim
30hx 2 + 2h 2 + h
h→0
h
e lim 5
a lim
d lim
Example
3
4 Find:
3x 2 h − 2xh 2 + h
h→0
h
PL
Example
E
2 A population of insects grows so that the population, P, at time t (days) is given by
P = 1000 + t2 + t, where t > 0. By finding the gradient of the chord between the points
where t = 3 and t = 3 + h, find an estimate for the rate of growth of the insect population
at time t = 3.
c lim 20 − 10h
h→0
h→0
(x + h)2 + 2(x + h) − (x 2 + 2x)
i.e. the derivative of y = x2 + 2x
h→0
h
(5 + h)2 + 3(5 + h) − 40
i.e. the gradient of y = x2 + 3x, where x = 5
b lim
h→0
h
(x + h)3 + 2(x + h)2 − (x 3 + 2x 2 )
i.e. the derivative of y = x3 + 2x2
c lim
h→0
h
a lim
Example
1
M
P1: FXS/ABE
5 For a curve with equation y = 3x2 − x:
SA
a find the gradient of chord PQ, where P is the point (1, 2) and Q is the point
((1 + h), 3(1 + h)2 − (1 + h))
b find the gradient of PQ when h = 0.1
c find the gradient of the curve at P.
2
:
x
a find
the gradient of chord AB where A is the point (2, 1) and B is the point
2
(2 + h),
2+h
6 For a curve with equation y =
b find the gradient of AB when h = 0.1
c find the gradient of the curve at A.
7 For a curve with equation y = x2 + 2x − 3:
a find the gradient of chord PQ, where P is the point (2, 5) and Q is the point ((2 + h),
(2 + h)2 + 2(2 + h) −3)
b find the gradient of PQ when h = 0.1
c find the gradient of the curve at P.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
542
19.2
Essential Mathematical Methods 1 & 2 CAS
The derived function
In Section 19.1 we saw how the gradient function of a function with rule f (x) could be derived
by considering the gradient of a chord PQ on the curve of y = f (x).
Consider the graph y = f (x) of the function f : R → R.
The gradient of the chord PQ =
f (x + h) − f (x)
x +h−x
=
f (x + h) − f (x)
h
Therefore the gradient of the graph at P is
given by
y = f (x)
Q
(x + h, f (x + h))
P (x, f (x))
0
PL
f (x + h) − f (x)
lim
h→0
h
y
E
P1: FXS/ABE
x
The reader is referred to Section 19.4 for a further discussion of limits.
The gradient or derived function is denoted by f ,
f (x + h) − f (x)
where f : R → R and f (x) = lim
h→0
h
M
In this chapter only polynomial functions are considered. For a polynomial function the
derived function always exists and is defined for every number in the domain of f.
Determining the derivative of an expression or the derived function by evaluating the limit is
called differentiation by first principles.
Example 5
SA
For f (x) = x2 + 2x find f (x) by first principles.
Solution
f (x + h) − f (x)
h→0
h
2
(x + h) + 2(x + h) − (x 2 + 2x)
= lim
h→0
h
x 2 + 2xh + h 2 + 2x + 2h − x 2 − 2x
= lim
h→0
h
2
2xh + h + 2h
= lim
h→0
h
= lim 2x + h + 2
f (x) = lim
h→0
∴
= 2x + 2
f (x) = 2x + 2
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
543
Example 6
For f (x) = 2 − x3 find f (x) by first principles.
Solution
f (x + h) − f (x)
h
2 − (x + h)3 − (2 − x 3 )
= lim
h→0
h
3
2 − (x + 3x 2 h + 3xh 2 + h 3 ) − (2 − x 3 )
= lim
h→0
h
2
2
3
−3x h − 3xh − h
= lim
h→0
h
f (x) = lim
E
h→0
PL
P1: FXS/ABE
= lim −3x 2 − 3xh − h 2
h→0
= −3x 2
Using the TI-Nspire
SA
M
Define f (x) = 2 − x 3 .
Calculate the gradient of the chord,
f (x + h) − f (x)
.
h
Select Limit from the Calculus menu
(b 4 3) and complete as shown.
Using the Casio ClassPad
It is a good idea to select Edit—Clear all
variables at this point as we will be storing
information as variables and a clean variable list
is advisable.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
544
Essential Mathematical Methods 1 & 2 CAS
In
select Interactive—Define and enter
the information, f (x) = 2 − x 3 , as shown.
Note: Select f from abc keyboard and x from 2D
VAR keyboard.
PL
Enter the expression (f(x + h) − f(x))/ h as
shown and click EXE to produce the algebraic
expression.
Copy and paste the answer to the next entry
line, then highlight it and select
Interactive—Transformation—simplify to
produce the simplified expression.
E
P1: FXS/ABE
M
Copy and paste this answer to the next entry
line, then highlight it and select
Interactive—Calculation—lim and set the
variable as h, Point as 0 and Direction as 0.
The following results have been obtained:
For f (x) = x, f (x) = 1.
For f (x) = x2 , f (x) = 2x.
For f (x) = x3 , f (x) = 3x2 .
SA
For f (x) = x4 , f (x) = 4x3 .
For f (x) = 1, f (x) = 0.
This suggests the following general result:
For
f (x) = xn , f (x) = nx n−1 , n = 1, 2, 3, . . .
and for f (x) = 1,
f (x) = 0
From the previous section it can be seen that for k, a constant:
If f (x) = kx n , the derivative function f has rule f (x) = knx n−1 .
It is worth making a special note of the results:
If g(x) = k f (x), where k is a constant, then g (x) = k f (x).
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
545
That is, the derivative of a number multiple is the multiple of the derivative.
For example, for g(x) = 5x2 , the derived function g (x) = 5(2x) = 10x.
Another important rule for differentiation is:
If f (x) = g(x) + h(x), then f (x) = g (x) + h (x).
E
That is, the derivative of the sum is the sum of the derivatives.
For example, for f (x) = x2 + 2x, the derived function f (x) = 2x + 2.
The process of finding the derivative function is called differentiation.
Example 7
Find the derivative of x5 − 2x3 + 2, i.e. differentiate x5 − 2x3 + 2 with respect to x.
Solution
then
PL
P1: FXS/ABE
f (x) = x 5 − 2x 3 + 2
f (x) = 5x 4 − 2(3x 2 ) + 2(0)
= 5x 4 − 6x 2
Example 8
M
Find the derivative of f (x) = 3x3 − 6x2 + 1 and f (1).
Solution
SA
then
f (x) = 3x 3 − 6x 2 + 1
f (x) = 3(3x 2 ) − 6(2x) + 1(0)
= 9x 2 − 12x
f (1) = 9 − 12
= −3
Using the TI-Nspire
Solutions to Example 7 and 8
Select Derivative from the Calculus menu
(b 4 1) and complete as shown.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
546
Essential Mathematical Methods 1 & 2 CAS
PL
E
Define f (x) = 3x 3 − 6x 2 + 1. Select
Derivative from the Calculus menu (b 5
1) and complete as shown.
Using the Casio ClassPad
Solutions to Examples 7 and 8
Enter and highlight the expression, then from
and
the keyboard select 2D—Calc—
enter x as the variable.
SA
M
In
select Action—Command—Define
and enter f (x) = 3x 3 − 6x 2 + 1 as shown (f is
and x is from the
from the abc menu in
grey x key on the calculator keyboard).
Select Interactive—Calculation—diff and
enter the information shown. Then click EXE.
In Action—Command—Define define
function g(x) = 9x 2 − 12x, the derivative
function for f(x).
Now enter g(1).
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
547
Example 9
Find the gradient of the curve determined by the rule f (x) = 3x3 − 6x2 + 1 at the point
(1, −2).
Solution
E
Now f (x) = 9x2 − 12x and f (1) = 9 − 12 = −3.
The gradient of the curve is −3 at the point (1, −2).
An alternative notation for the derivative is the following:
dy
dy
, so that
= 3x 2 .
dx
dx
PL
If y = x 3 , then the derivative can be denoted by
dy
and
dx
with the use of different symbols z, where z is a function of t. The derivative of z with respect
dz
.
to t is
y
dt
In this notation d is not a factor and cannot be cancelled.
P
This came about because in the eighteenth century the
δy
standard diagram for finding the limiting gradient was
Q
labelled as in the figure. ( is the lower case Greek letter
δx
for ‘d’, and is pronounced delta.)
x
x means a difference in x.
0
y means a difference in y.
M
In general, if y is a function of x, the derivative of y with respect to x is denoted by
SA
Example 10
a If y = t 2 , find
dy
.
dt
b If x = t 3 + t, find
dx
.
dt
c If z =
1 3
dz
x + x 2 , find
.
3
dx
Solution
a
y = t2
dy
= 2t
then
dt
b
x = t3 + t
dx
= 3t 2 + 1
then
dt
z=
c
then
1 3
x + x2
3
dz
= x 2 + 2x
dx
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
548
Essential Mathematical Methods 1 & 2 CAS
Example 11
dy
.
dx
dy
x 2 + 3x
, find
.
c For y =
x
dx
Solution
b For z = (2t − 1)2 (t + 2), find
d
a It is first necessary to write
y = (x + 3)2 in expanded form.
∴
Differentiate y = 2x3 − 1 with respect to x.
b Expanding:
z = (4t 2 − 4t + 1)(t + 2)
y = x 2 + 6x + 9
dy
= 2x + 6
dx
and
= 4t 3 − 4t 2 + t + 8t 2 − 8t + 2
= 4t 3 + 4t 2 − 7t + 2
and
c First divide by x:
∴
dz
.
dt
E
a For y = (x + 3)2 , find
dz
= 12t 2 + 8t − 7
dt
PL
P1: FXS/ABE
d
∴
y = x +3
dy
=1
dx
y = 2x 3 − 1
dy
= 6x 2
dx
Operator notation
M
d
‘Find the derivative of 2x2 − 4x with respect to x’ can also be written as
(2x 2 − 4x), and, in
dx
d
( f (x)) = f (x).
general,
dx
Example 12
SA
Find:
d
a
(5x − 4x 3 )
dx
b
d
(5z 2 − 4z)
dz
c
d
(6z 3 − 4z 2 )
dz
Solution
a
d
(5x − 4x 3 )
dx
= 5 − 12x 2
b
d
(5z 2 − 4z)
dz
= 10z − 4
c
d
(6z 3 − 4z 2 )
dz
= 18z 2 − 8z
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
549
Example 13
Find the coordinates of the points on curves determined by each of the following equations at
which the gradient has the given value:
a y = x3 ; gradient = 8
b y = x2 − 4x + 2; gradient = 0
c y = 4 − x3 ; gradient = −6
a y = x 3 implies
∴ 3x 2 = 8
E
Solution
dy
= 3x 2
dx
b y = x 2 − 4x + 2 implies
∴ 2x − 4 = 0
√
dy
= 2x − 4
dx
∴
x =2
±2 6
8
=
∴ coordinates are (2, −2)
3
3
∴ coordinates are
√
√
√ √ −2 6 −16 6
2 6 16 6
,
and
,
3
9
3
9
∴
x =±
PL
P1: FXS/ABE
c y = 4 − x 3 implies
∴ −3x 2 = −6
∴
x2 = 2
√
x = ± 2.
M
∴
dy
= −3x 2
dx
∴ coordinates are
1
3
22 , 4 − 22
1
3
and −2 2 , 4 + 2 2
Using the TI-Nspire
SA
Define f (x) = 4 − x 3 .
Take the Derivative and store as d f (x).
Solve the equation d f (x) = −6.
Substitute in f (x) to find the y-coordinates.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
550
Essential Mathematical Methods 1 & 2 CAS
Using the Casio ClassPad
E
In Interactive—Define, define the function
f (x) = 4 − x 3 .
Select Interactive—
Calculate—diff and
enter as shown.
PL
Enter and highlight the equation −3x 2 = −6
for x. (You can copy the answer from the line
above to save re-typing.)
√
√
Now enter f ( 2) and f (− 2) to find the
required y-values.
Exercise 19B
5, 6
1 For each of the following, find f (x) by finding lim
M
Examples
a
e
Example
7
f (x) = 3x2
b
f (x) = 2x3 − 4
h→0
f (x + h) − f (x)
:
h
f (x) = 4x
c f (x) = 3
f f (x) = 4x2 − 5x
f (x) = 3x2 + 4x + 3
f (x) = 3 − 2x + x2
d
g
2 Find the derivative of each of the following with respect to x.
SA
a x2 + 4x
d
1 2
x − 3x + 4
2
b 2x + 1
c x3 − x
e 5x3 + 3x2
f −x 3 + 2x 2
3 For each of the following find f (x):
Example
10
a
f (x) = x12
b
f (x) = 3x7
c
f (x) = 5x
d
f (x) = 5x + 3
e
f (x) = 3
f
f (x) = 5x2 − 3x
g
f (x) = 10x5 + 3x4
h
1
1
f (x) = 2x 4 − x 3 − x 2 + 2
3
4
dy
:
dx
b y = 10
4 For each of the following, find
a y = −x
1
d y = (x 3 − 3x + 6)
e y = (x + 1)(x + 2)
3
10x 5 + 3x 4
, x = 0
g y=
2x 2
c y = 4x3 − 3x + 2
f y = 2x(3x2 − 4)
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
Example
9
551
5 a For the curve with equation y = x3 + 1 find the gradient at points:
i (1, 2)
ii (a, a 3 + 1)
b Find the derivative of x3 + 1 with respect to x.
dy
dy
. Hence show that
≥ 0 for all x, and
6 a Given that y = x3 − 3x2 + 3x, find
dx
dx
interpret this in terms of the graph of y = x3 − 3x2 + 3x.
dy
x 2 + 2x
, for x = 0, find
.
x
dx
c Differentiate y = (3x + 1)2 with respect to x.
E
b Given that y =
7 At the points on the following curves corresponding to the given values of x, find the
y-coordinate and the gradient.
b y = x2 + x + 1, x = 0
PL
a y = x2 − 2x + 1, x = 2
c y = x2 − 2x, x = −1
d y = (x + 2)(x − 4), x = 3
1
e y = 3x2 − 2x3 , x = −2
f y = (4x − 5)2 , x =
2
8 a For each of the following, find f (x) and f (1), if y = f (x) then find the
{(x, y): f (x) = 1} i.e. the coordinates of the points where the gradient is 1.
1
1
ii y = 1 + x + x 2
i y = 2x2 − x
2
3
3
4
iii y = x + x
iv y = x − 31x
Example
12
M
b What is the interpretation of {(x, y): f (x) = 1} in terms of the graphs?
9 Find:
d
a
(3t 2 − 4t)
dt
d
(3y 2 − y 3 )
dy
SA
d
Example
13
b
d
(4 − x 2 + x 3 )
dx
c
d
(5 − 2z 2 − z 4 )
dz
e
d
(2x 3 − 4x 2 )
dx
f
d
(9.8t 2 − 2t)
dt
10 Find the coordinates of the points on the curves given by the following equations at which
the gradient has the given values:
a y = x2 ; gradient = 8
b y = x3 ; gradient = 12
c y = x(2 − x); gradient = 2
d y = x2 − 3x + 1; gradient = 0
e y = x3 − 6x2 + 4; gradient = −12
f y = x2 − x3 ; gradient = −1
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
552
19.3
Essential Mathematical Methods 1 & 2 CAS
Graphs of the derived or gradient function
y
Consider the gradient for different intervals
of the graph of y = g(x) shown opposite.
At a point (a, g(a)) of the graph y = g(x)
the gradient is g (a).
R(b, g (b))
a
b
x
0
E
Some of the features of the graph are:
x < b the gradient is positive,
i.e. g (x) > 0
x = b the gradient is zero, i.e. g (b) = 0
b < x < a the gradient is negative, i.e. g (x) < 0
x = a the gradient is zero, i.e. g (a) = 0
x > a the gradient is positive, i.e. g (x) > 0
y = g (x)
Example 14
For the graph of f : R → R, find:
b {x: f (x) < 0}
a {x: f (x) > 0}
Solution
S(a, g(a))
PL
P1: FXS/ABE
y
c {x: f (x) = 0}
a {x: f (x) > 0} = {x: −1 < x < 5} = (−1, 5)
(5, 6)
x
0
M
b {x: f (x) < 0} = {x: x < −1} ∪ {x : x > 5}
= (−∞, −1) ∪ (5, ∞)
c {x: f (x) = 0} = {−1, 5}
y = f (x)
(–1, –7)
Example 15
SA
Sketch the graph of y = f (x) for each of the following. (It is impossible to determine all
features.)
y
c
y
b
y
a
(–1.5, 4)
y = f (x)
y = f (x)
y = f (x)
0
2
x
1
x
x
–1
–3 –1
0
4
0
4
(3, –1)
(1, –4)
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
553
Solution
a
f (x) > 0 for x > 3
f (x) < 0 for x < 3
f (x) = 0 for x = 3
y
b f (x) = 1 for all x
y
y = f' (x)
1
0
3
y = f ′(x)
x
0
f
f
f
f
y
(x) > 0 for x > 1
(x) < 0 for −1.5 < x < 1
(x) > 0 for x < −1.5
(−1.5) = 0 and f (1) = 0
y = f ′(x)
−1.5
0
x
1
PL
c
x
E
P1: FXS/ABE
An angle associated with the gradient
of a curve at a point
M
The gradient of a curve at a point is the gradient of the tangent at that point. A straight line, the
tangent, is associated with each point on the curve.
If is the angle a straight line makes with the positive direction of the x-axis, then the
gradient, m, of the straight line is equal to tan , i.e., tan = m.
If = 45◦ then tan = 1 and the gradient is 1.
If = 20◦ then the gradient of the straight line is tan 20◦ .
If = 135◦ then tan = −1 and the gradient is −1.
Example 16
SA
Find the coordinates of the points on the curve with equation y = x2 − 7x + 8 at which the
tangent:
a makes an angle of 45◦ with the positive direction of the x-axis
b is parallel to the line y = −2x + 6.
Solution
a
dy
= 2x − 7
dx
2x − 7 = 1 (tan 45◦ = 1)
and 2x = 8
∴
x =4
y = 42 − 7 × 4 + 8 = −4
coordinates are (4, −4)
b
Note: y = −2x + 6 has gradient −2
∴
2x − 7 = −2
and 2x = 5
5
∴ x=
2
coordinates are
5 13
,−
2
4
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
554
Essential Mathematical Methods 1 & 2 CAS
Example 17
E
The planned path for a flying saucer leaving a planet is defined by the equation
1
2
y = x 4 + x 3 for x > 0.
4
3
The units are kilometres. (The x-axis is horizontal and the y-axis vertical.)
a Find the direction of motion when the x-value is:
i 2
ii 3
b Find a point on the flying saucer’s path where the path is inclined at 45◦ to the positive
x-axis. (i.e. where the gradient of the path is 1).
c Are there any other points on the path which satisfy the situation described in part b?
Solution
a
dy
= x 3 + 2x 2
dx
PL
P1: FXS/ABE
dy
= 8 + 8 = 16
dx
tan−1 16 = 86.42◦ (to the x-axis)
i When x = 2,
dy
= 27 + 18 = 45
dx
tan−1 45 = 88.73◦ (to the x-axis)
ii When x = 3,
SA
M
b, c When the flying saucer is flying at 45◦ to the direction of the x-axis, the gradient
of the curve of its path is given by tan 45◦ .
dy
= tan 45◦ .
Thus to find the point at which this happens we consider equation
dx
∴
x 3 + 2x 2 = 1
3
∴
x + 2x 2 − 1 = 0
∴ (x + 1)(x 2 + x − 1) = 0
√
−1 ± 5
∴
x = −1 or x =
2
√
−1 + 5
The only acceptable solution is x =
(x ≈ 0.62) as the other two
2
possibilities give negative values for x and we are only considering positive
values for x.
Exercise 19C
1 On which of the following curves is
y
a
0
dy
positive for all values of x?
dx
y
b
x
0
c
y
x
0
x
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
d
y
y
e
x
0
x
0
2 On which of the following curves is
a
x
y
x
0
PL
0
c
dy
negative for all values of x?
dx
y
b
E
y
y
d
x
0
x
M
0
y
SA
e
555
y
f
x
0
x
0
3 For the function f (x) = 2(x − 1)2 find the values of x for which:
a
d
Example
15
f (x) = 0
f (x) < 0
b
e
f (x) = 0
f (x) = −2
4 For the graph of y = h(x) shown here find:
a {x: h (x) > 0}
b {x: h (x) < 0}
c {x: h (x) = 0}
c
f (x) > 0
y
–3,
5 12
(4, 6)
x
0
1
, –3
2
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
556
Essential Mathematical Methods 1 & 2 CAS
5 Which of the graphs labelled A–F correspond to each of the graphs labelled a–f?
a
y
A
y
d
D
dy
dy
dx
dx
0
x
y
b
0
y
e
B
x
y
c
x
E
dy
dx
dy
dx
PL
x
0
0
E
x
0
x
y
f
x
0
0
dy
dx
0
Example
15
x
x
0
dy
dx
x
0
y
6 For the graph of y = f (x) find:
SA
16
x
0
a {x: f (x) > 0}
b {x: f (x) < 0}
c {x: f (x) = 0}
Example
x
0
F
C
M
P1: FXS/ABE
(1.5, 3)
x
0
(–1, –2)
7 Sketch the graph of y = f (x) for each of the following
a y
y
b
y
c
(3, 4)
0 1
5
y
d
(3, 4)
y = f (x)
1
x
0 1
x
0
3
x
y = f (x)
0
x
(–1, –3)
Example
17
8 Find the coordinates of the points on the curve y = x2 − 5x + 6 at which the tangent:
a makes an angle of 45◦ with the positive direction of the x-axis, i.e. where the gradient
is 1
b is parallel to the line y = 3x + 4.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
Example
17
557
9 Find the coordinates of the points on the parabola y = x2 − x − 6 at which:
a the gradient is zero
b the tangent is parallel to the line x + y = 6.
10 Use a calculator to plot the graph of y = f (x) where:
a
f (x) = sin x
b
f (x) = cos x
c
f (x) = 2x
a Find its speed after t seconds.
E
11 A car moves away from a set of traffic lights so that the distance S(t) metres covered after
t seconds is modelled by S(t) = (0.2)t 3 .
b What will its speed be when t = 1, 3, 5?
PL
12 A rocket is launched from Cape York Peninsula so that after t seconds its height
h(t) metres is given by h(t) = 20t2 , 0 ≤ t ≤ 150. After 2 12 minutes this model is no
longer appropriate.
a Find the height and the speed of the rocket when t = 150.
b After how long will its speed be 1000 m/s?
13 The curve with equation y = ax2 + bx has a gradient of 3 at the point (2, −2).
a Find the values of a and b.
b Find the coordinates of the point where the gradient is 0.
19.4
Limits and continuity
M
Limits
SA
We consider the limit of a function f (x) to be the value that f (x) approaches as x approaches a
given value. lim f (x) = p means that, as x approaches a, f (x) approaches p. It is important to
x→a
understand that it is possible to get as close as desired to p as x approaches a. Note that f (x)
may or may not be defined at x = a.
For many functions f (a) is defined, so to evaluate the limit we simply substitute the value a
into the rule for the function.
Example 18
If f (x) = 3x2 find lim 3x 2 .
x→2
Solution
Since f (x) = 3x2 is defined at x = 2
lim 3x 2 = 3(2)2
= 12
x→2
If the function is not defined at the value for which the limit is to be found, a different
procedure is used.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
558
Essential Mathematical Methods 1 & 2 CAS
Example 19
For f (x) =
2x 2 − 5x + 2
, x = 2, find lim f (x).
x→2
x −2
Solution
Observe that f (x) is defined for x ∈ R \ {2}.
Examine the behaviour of f (x) for values of x ‘near’ 2.
x>2
f (1.7) = 2.4
f (1.8) = 2.6
f (1.9) = 2.8
f (1.99) = 2.98
f (1.999) = 2.998
f (2.3) = 3.6
f (2.2) = 3.4
f (2.1) = 3.2
f (2.01) = 3.02
f (2.001) = 3.002
y
PL
x<2
E
P1: FXS/ABE
0
f
x
1
2
–1
M
From the table it is apparent that as x takes values
closer and closer to 2, regardless of whether
x approaches 2 from the left or from the right, the
values of f (x) become closer and closer to 3.
i.e. lim f (x) = 3
3
x→2
This may also be seen by observing that:
(2x − 1)(x − 2)
, n = 2
x −2
= 2x − 1, x = 2,
f (x) =
SA
The graph of f : R \ {2} → R, f (x) = 2x − 1 is shown.
The following important results are useful for the evaluation of limits:
lim ( f (x) + g(x)) = lim f (x) + lim g(x)
x→c
x→c
x→c
i.e. the limit of the sum is the sum of the limits.
lim k f (x) = k lim f (x), k being a given number
x→c
x→c
lim ( f (x)g(x)) = lim f (x) lim g(x)
x→c
x→c
x→c
i.e. the limit of the product is the product of the limits.
lim f (x)
f (x)
x→c
=
, provided lim g(x) = 0
lim
x→c g(x)
x→c
lim g(x)
x→c
i.e. the limit of the quotient is the quotient of the limits.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
559
Example 20
Find:
a lim (x 2 + 2)
b lim
x 2 − 3x
x −3
d lim (2x + 1)(3x − 2)
e lim
x 2 − 7x + 10
x 2 − 25
x→0
x→3
x→3
x→3
c lim
x→2
x2 − x − 2
x −2
E
Solution
a lim (x 2 + 2) = lim x 2 + lim 2 = 0 + 2 = 2
x→0
x→0
x→0
x 2 − 3x
x(x − 3)
b lim
= lim
= lim x = 3
x→3 x − 3
x→3 x − 3
x→3
x2 − x − 2
(x − 2)(x + 1)
c lim
= lim
= lim (x + 1) = 3
x→2
x→2
x→2
x −2
x −2
d lim (2x + 1)(3x − 2) = lim (2x + 1) lim (3x − 2) = 7 × 7 = 49
x→3
PL
P1: FXS/ABE
x→3
x→3
lim (x − 2)
x 2 − 7x + 10
(x − 2)(x − 5)
1
x→3
=
lim
=
=
e lim
2
x→3
x→3 (x + 5)(x − 5)
x − 25
lim (x + 5)
8
x→3
Example 21
Find:
a lim (3h + 4)
b lim 4x(x + 2)
x→2
M
h→0
5x + 2
x→3 x − 2
c lim
Solution
a lim (3h + 4) = lim (3h) + lim (4)
h→0
h→0
h→0
=0+4
=4
SA
b lim 4x(x + 2) = lim (4x) lim (x + 2)
x→2
c lim
x→3
x→2
=8×4
= 32
x→2
5x + 2
= lim (5x + 2) ÷ lim (x − 2)
x→3
x→3
x −2
= 17 ÷ 1
= 17
The notation of limits is used to describe the behaviour of graphs, and a similar notation has
been used previously in the book.
1
Consider f : R \ {0} → R, f (x) = 2 . Observe that as x → 0, both from the left and from
x
the right, f (x) increases without bound. The limit notation for this is lim f (x) = ∞.
x→0
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
560
Essential Mathematical Methods 1 & 2 CAS
1
For g: R \ {0} → R, g(x) = , the behaviour of g(x) as x approaches 0 from the left is
x
different from the behaviour as x approaches 0 from the right.
With limit notation this is written as:
lim g(x) = −∞
x→0−
lim f (x) = ∞
and
x→0+
y
E
Now examine this function as the magnitude of x becomes
very large. It can be seen that as x increases without
bound through positive values, the corresponding values
of g(x) approach zero. Likewise as x decreases
without bound through negative values, the
corresponding values of g(x) also approach zero.
Symbolically this is written as:
lim g(x) = 0
x→∞
1
g(x) = x
0
x
PL
P1: FXS/ABE
and
lim g(x) = 0
x→−∞
Many functions approach a limiting value or limit as x approaches ±∞.
Left and right limits
An idea which is useful in the following discussion is the existence of limits from the left and
from the right.
If the value of f (x) approaches the number p as x approaches a from the right-hand side,
then it is written as lim+ f (x) = p. If the value of f (x) approaches the number p as x
M
x→a
approaches a from the left-hand side, then it is written as lim− f (x) = p.
x→a
SA
The limit as x approaches a exists only if the limits from the left and the right both exist and
are equal. Then lim f (x) = p.
x→a
The following is an example of when the limit does not exist for a particular value.
 3
 x for 0 ≤ x < 1
Let f (x) = 5 for x = 1

6 for 1 < x ≤ 2
It is clear from the graph of f that the
lim f (x) does not exist. However, if x is
x→1
allowed to approach 1 from the left, then
f (x) approaches 1. On the other hand if x is
allowed to approach 1 from the right, then
f (x) approaches 6. Also note that f (1) = 5.
y
6
5
4
3
2
1
0
x
1
2
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
561
Continuity at a point: informal definition
A function with rule f (x) is said to be continuous when x = a if the graph of y = f (x) can be
drawn through the point with coordinates (a, f (a)) without a break. Otherwise there is said to
be a discontinuity at x = a.
Most of the functions considered in this course are continuous for their domains.
A more formal definition of continuity follows.
A function f is continuous at a point a if f (a), lim+ f (x) and lim− f (x) all exist and
x→a
x→a
E
are equal. Or equivalently:
A function f is continuous at the point x = a if the following three conditions are met:
1 f (x) is defined at x = a
2 lim f (x) exists
3 lim f (x) = f (a)
x→a
x→a
The function is said to be discontinuous at a point if it is not continuous at that point. A
function is said to be continuous everywhere if it is continuous for all real numbers.
1
The polynomial functions are all continuous for R. The function with rule f (x) = has a
x
discontinuity at x = 0, as f (0) is not defined. It is continuous elsewhere in its domain.
Hybrid functions, as introduced in Chapter 6, provide examples of functions which have
points of discontinuity where the function is defined.
PL
P1: FXS/ABE
M
Example 22
State the values for x for which the functions whose graphs are shown below have a
discontinuity.
y
c
y
a
b
y
3
SA
2
0
3
2
2
1
x
1
–1
3
–1
0
1
1
x
x
–1
1
0
1
Solution
a There is a discontinuity at x = 1, as f (1) = 3 but lim+ f (x) = lim− f (x) = 2.
x→1
x→1
b There is a discontinuity at x = −1, as f (−1) = 2 and lim − f (x) = 2 but
x→−1
lim + f (x)= −∞ and a discontinuity at x = 1 as f (1) = 2 and lim− f (x) = 2
x→−1
x→1
but lim+ f (x) = 3.
x→1
c There is a discontinuity at x = 1, as f (1) = 1 and lim− f (x) = 1 but
lim+ f (x) = 2.
x→1
x→1
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
562
Essential Mathematical Methods 1 & 2 CAS
Example 23
For each of the following functions state the values of x for which there is a discontinuity and
use the definition of continuity in terms of f (a), lim+ f (x) and lim− f (x) to explain why each
x→a
Solution
x→a
if x ≥ 0
if x < 0
f (x) =
b
E
if x ≤ −1
if −1 < x < 0
if x ≥ 0
if x ≥ 0
if x < 0
x2
−2x + 1
f (x) =
d
if x ≥ 0
if x < 0
x2 + 1
−2x + 1
if x ≥ 0
if x < 0
PL
is discontinuous:
2x
a f (x) =
−2x + 1


x
c f (x) = x 2


−2x + 1
x
e f (x) =
−2x
a f (0) = 0 but lim− f (x) = 1, therefore there is a discontinuity at x = 0
x→0
b f (0) = 0 but lim− f (x) = 1, therefore there is a discontinuity at x = 0
x→0
c f (−1) = −1 but lim + f (x) = 1, therefore there is a discontinuity at x = 1
x→−1
f (0) = 1 but lim− f (x) = 0, therefore there is a discontinuity at x = 0
x→0
e No discontinuity
M
d No discontinuity
Exercise 19D
Examples
18–21
1 Find the following limits:
b lim (x − 5)
a lim 15
SA
x→3
d
(t − 2)
t→−3 (t + 5)
lim
t2 − 1
t→1 t − 1
x3 − 8
j lim
x→2 x − 2
g lim
Example
22
c
x→6
t 2 + 2t + 1
t→−1
t +1
h lim
x→9
√
1
2
(x + 2)2 − 4
x→0
x
lim
e
lim (3x − 5)
x→
f lim
x 2 − 2x
x→0
x
2
x − 3x + 2
l lim 2
x→1 x − 6x + 5
x +3
i lim
3x 2 − x − 10
x→2 x 2 + 5x − 14
k lim
2 For each of the following graphs give the values of x at which a discontinuity occurs. Give
reasons.
a
y
y
b
6
2
x
1
3 4
x
2
5 7
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
Example
23
563
3 For each of the following functions state the values of x at which there is a discontinuity
and use the definition of continuity in terms of f (a), lim+ f (x) and lim− f (x) to explain
x→a
x→a
E
why each stated value of x corresponds to a discontinuity.
3x
if x ≥ 0
x 2 + 2 if x ≥ 1
a f (x) =
b f (x) =
−2x + 2 if x < 0
−2x + 1 if x < 1


if x ≤ −1
−x
c f (x) = x 2
if −1 < x < 0


−3x + 1 if x ≥ 0
19.5
PL
4 The rule of a particular function is given below. For what values of x is the graph of this
function discontinuous?


x <1
2,
y = (x − 4)2 − 9, 1 ≤ x < 7


x − 7,
x ≥7
When is a function differentiable?
f (x + h) − f (x)
exists.
h
The polynomial functions considered in this chapter are differentiable for all x. However
this is not true for all functions.
x
if x ≥ 0
Let f : R → R, f (x) =
−x if x < 0
A function f is said to be differentiable at x if lim
M
h→0
SA
This function is called the modulus function or absolute value function and it is denoted by
f (x) = |x|, f is not differentiable at x = 0:
h
y

h>0

f (0 + h) − f (0)
h
=
−h
y = x

h

h<0
h
1
h>0
=
−1 h < 0
f (0 + h) − f (0)
does not exist.
h→0
h
i.e. f is not differentiable at x = 0.
So lim
0
x
Note: The gradient to the left of 0 is −1 and to the right of 0 the gradient is 1.
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
564
Essential Mathematical Methods 1 & 2 CAS
Example 24
x
if x ≥ 0
(= |x|)
−x if x < 0
Sketch the graph of the derivative for a suitable domain.
y
Let f : R → R, f (x) =
Solution 1
if x > 0
f (x) =
−1 if x < 0
f (x) is not defined at x = 0
1
x
0
E
–1
y
Example 25
2
Draw a sketch graph of f where the graph of f is as illustrated.
Indicate where f is not defined.
Solution
PL
P1: FXS/ABE
–1
0
x
1
y
The derivative does not exist at x = 0;
i.e. the function is not differentiable at x = 0.
2
–1
M
0
–2
1
x
SA
It was shown in the previous section that some hybrid functions are continuous for R. There
are hybrid functions which are differentiable for R. The smoothness of the ‘joins’ determines if
this is the case.
Example 26
For the function with following rule find f (x) and sketch the graph of y = f (x):
x 2 + 2x + 1 if x ≥ 0
f (x) =
2x + 1
if x < 0
Solution
f (x) =
y
2x + 2
2
if x ≥ 0
if x < 0
In particular f (0) is defined and is equal
to 2. Also f (0) = 1. The two sections of
the graph of y = f (x) join smoothly at (0, 1).
2
0
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
x
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
565
Chapter 19 — Differentiation of Polynomials
Example 27
For the function with rule
x 2 + 2x + 1
f (x) =
x +1
if x ≥ 0
if x < 0
Solution
f (x) =
E
state the set of values for which the derivative is defined, find f (x) for this set of values and
sketch the graph of y = f (x).
y
2x + 2
1
if x > 0
if x < 0
2
PL
f (0) is not defined as the limits from
1
the left and right are not equal. The
function is differentiable for R\{0}.
x
0
Exercise 19E
24–26
1 In each of the figures below a function graph f is given. Sketch the graph of f . Obviously
your sketch of f cannot be exact, but f (x) should be 0 at values of x for which the
gradient of f is zero; f (x) should be < 0 where the original graph slopes downward, and
so on.
M
Examples
y
a
f
0
d
f
–1
2 4
y
e
y
x
–4 –2 0
1
SA
–1
x
y
c
y
b
x
0
1
f
y
f
f
–1
Example
26
0
1
f
x
–1
0
f
x
1
–1
0
x
1
2 For the function with following rule find f (x) and sketch the graph of y = f (x):
−x 2 + 3x + 1
if x ≥ 0
f (x) =
3x + 1
if x < 0
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
566
Essential Mathematical Methods 1 & 2 CAS
27
3 For the function with following rule state the set of values for which the derivative is
defined, find f (x) for this set of values and sketch the graph of y = f (x):
x 2 + 2x + 1
if x ≥ 1
f (x) =
−2x + 3
if x < 1
Example
27
4 For the function with following rule state the set of values for which the derivative is
defined, find f (x) for this set of values and sketch the graph of y = f (x):
if x ≥ −1
−x 2 − 3x + 1
f (x) =
−2x + 3
if x < −1
SA
M
PL
E
Example
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
567
The notation for limit as h approaches 0 is written as lim .
h→0
The following are important results for limits:
r lim ( f (x) + g(x)) = lim f (x) + lim g(x)
x→c
x→c
x→c
r
i.e. the limit of the sum is the sum of the limits.
lim k f (x) = k lim f (x), k being a given number
r
lim ( f (x)g(x)) = lim f (x) lim g(x)
x→c
x→c
x→c
E
r
x→c
x→c
i.e. the limit of the product is the product of the limits.
lim f (x)
f (x)
x→c
=
, provided lim g(x) = 0
lim
x→c g(x)
x→c
lim g(x)
x→c
i.e. the limit of the quotient is the quotient of the limits.
A function f is defined as continuous at the point x = a if three conditions are met:
1 f (x) is defined at x = a
2
3
lim f (x) exists
x→a
lim f (x) = f (a)
x→a
M
The function is said to be discontinuous at a point if it is not continuous at that point. We
say that a function is continuous everywhere if it is continuous for all real numbers.
y
For the graph of y = f (x) of the function f : R → R:
f (x + h) − f (x)
The gradient of the chord PQ =
.
h
Q (x + h, f (x+h))
The gradient of the graph at P is given by
f (x + h) − f (x)
P (x, f (x))
lim
.
h→0
h
x
0
SA
This limit gives a rule for the derived function denoted by f , where f : R → R and
f (x + h) − f (x)
h
The general rule of the derived function of f (x) = xn , n = 1, 2, 3, . . .
For f (x) = xn , f x = nx n−1 , n = 1, 2, 3, . . .
f (x) = lim
h→0
For f (x) = 1, f (x) = 0
For example:
For f (x) = x2 , f (x) = 2x.
For f (x) = x3 , f (x) = 3x2 .
For f (x) = x4 , f (x) = 4x3 .
For f (x) = 1, f (x) = 0.
The derivative of a number multiple is the multiple of the derivative:
For g(x) = k f (x), where k is a constant
g (x) = k f (x)
For example:
g(x) = 3x2 , g (x) = 3(2x) = 6x
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Review
Chapter summary
PL
P1: FXS/ABE
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Essential Mathematical Methods 1 & 2 CAS
The derivative of a constant is always zero:
For g(x) = a, g (x) = 0
E
For example: f (x) = 27.3, f (x) = 0
The derivative of the sum is the sum of the derivatives:
If f (x) = g(x) + h(x) then
f (x) = g (x) + h (x).
For example: f (x) = x 2 + x 3 , f (x) = 2x + 3x 2
g(x) = 3x 2 + 43 , g (x) = 3(2x) + 4(3x 2 ) = 6x + 12x 2
y
At a point (a, g(a)) on the curve y = g(x) the gradient is g (a).
For x < b the gradient is positive, i.e. g (x) > 0
R(b, g(b))
x = b the gradient is zero, i.e. g (b) = 0
y = g (x)
PL
b < x < a the gradient is negative, i.e. g (a) < 0
x = a the gradient is zero, i.e. g (a) = 0
b
a
0
x > a the gradient is positive, i.e. g (x) > 0
S(a, g(a))
Multiple-choice questions
M
1 The gradient of the curve y = x 3 + 4x at the point where x = 2 is
A 12
B 4
C 10
D 16
E 8
2 The gradient of the chord of the curve y = 2x between the points where x = 1 and
x = 1 + h is given by
B 4 + 2h
C 4
D 4x
E 4+h
A 2(x + h)2 − 2x2
dy
4
3
3 If y = 2x − 5x + 2, then
equals
dx
B 4x 4 − 15x 2 + 2
C 4x 4 − 10x 2
A 8x 3 − 5x 2 + 2
3
3
2
D 8x − 15x + 2
E 8x − 15x
SA
Review
568
2
4 If f (x) = x 2 (x + 1) then f (−1) equals
A −1
B 1
C 2
6 If y =
D −2
E 5
D 2x + 9
E 2x
5 If f (x) = (x − 3) , then f (x) equals
A x −3
B x −6
C 2x − 6
2
dy
2x + 9x
, then
equals
3x
dx
4
2
2x 4
8x 3 + 18x
+ 6x
B 2x + 3
C 2x 2 + 3
D
3
3
dy
7 Given that y = x2 − 6x + 9, the values of x for which
≥ 0 are
dx
A x ≥3
B x >3
C x ≥ −3
D x ≤ −3
A
E 8x 3 + 18x
E x <3
8 If y = 2x − 36x , the points at which the tangent to the curve is parallel to the x-axis are
A 1, 0 and 3
B 0 and 3
C −3 and 3
D 0 and −3
E −3, 0 and 3
4
2
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
x
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Chapter 19 — Differentiation of Polynomials
569
E
Short-answer questions (technology-free)
dy
when:
dx
a y = 3x2 − 2x + 6
d y = 4(2x − 1)(5x + 2)
b y=5
e y = (x + 1)(3x − 2)
c y = 2x(2 − x)
f y = (x + 1)(2 − 3x)
dy
when:
dx
a y = −x
b y = 10
c y=
1 Find
2 Find
(x + 3)(2x + 1)
4
x 4 + 3x 2
2x 3 − x 2
e y=
2x 2
3x
3 For each of the following functions find the y-coordinates and the gradient at the point on
the curve for the given value of x:
b y = x 2 − 2x, x = −1
a y = x 2 − 2x + 1, x = 2
c y = (x + 2)(x − 4), x = 3
d y = 3x 2 − 2x 3 , x = −2
M
d y=
SA
4 Find the coordinates of the points on the curves given by the following equations at which
the gradient has the given value:
dy
dy
=0
b y = x 3 − 6x 2 + 4;
a y = x 2 − 3x + 1;
= −12
dx
dx
dy
dy
= −1
d y = x 3 − 2x + 7;
c y = x 2 − x 3;
=1
dx
dx
dy
dy
=0
f y = x(x − 3)2 ;
e y = x 4 − 2x 3 + 1;
=0
dx
dx
5 For the function with rule f (x) = 3(2x −1)2 find the values of x for which:
c f (x) > 0
a f (x) = 0
b f (x) = 0
e f (x) > 0
f f (x) = 3
d f (x) < 0
6 The curve with equation y = ax2 + bx has a gradient of 3 at the point (1, 1). Find:
a the values of a and b
b the coordinates of the points where the gradient is 0.
7 Sketch the graph of y = f (x). (All details cannot be
determined but the axis intercepts and shape of graph
can be determined.)
y
5
–1
–1
2
x
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Review
9 The coordinates of the point on the graph of y = x 2 + 6x − 5 at which the tangent is
parallel to the line y = 4x are
A (−1, − 10) B (−1, −2)
C (1, 2)
D (−1, 4)
E (−1, 10)
dy
is equals
10 If y = −2x 3 + 3x 2 − x + 1, then
dx
B −6x 2 + 6x
C −6x 2 + 3x − 1
A 6x 2 + 6x − 1
E 6x 2 − 6x − 1
D −6x 2 + 6x − 1
PL
P1: FXS/ABE
P1: FXS/ABE
P2: FXS
9780521740524c19.xml
CUAU021-EVANS
August 23, 2008
14:3
Back to Menu >>>
Essential Mathematical Methods 1 & 2 CAS
y
8 For the graph of y = h(x) find:
a {x: h (x) > 0}
b {x: h (x) < 0}
c {x: h (x) = 0}
(4, 3)
y = h(x)
x
0
(–1, –4)
E
Extended-response questions
PL
1 The diagram to the right shows part of the graph
dy
against x.
of
dy
dx
dx
Sketch a possible shape of y against x over
the same interval if:
r y = −1 when x = −1
–1
0
r y =0
when x = 0
r y =1
when x = 2.
M
2 The graph shown is that of a polynomial of the form
P(x) = ax3 + bx2 + cx + d.
Find the values of a, b, c and d.
Note: Q(1, −2) is not a turning point.
2
5
x
y
R (–2, 3)
x
Q (1, –2)
1 5 1 4
x + x , x ≥ 0.
5
2
Units are in kilometres and x and y are the horizontal and vertical axes respectively.
a What will be the direction of motion (give the answer as angle between direction of
motion and the x-axis) when the x-value is
i 1 km?
ii 3 km?
b Find a value of x for which the gradient of the path is 32.
3 A body moves in a path described by the equation y =
SA
Review
570
4 A trail over a mountain pass can be modelled by the curve with equation
y = 2 + 0.12x − 0.01x3 , where x and y are, respectively, the horizontal and vertical
distances measured in kilometres, 0 ≤ x ≤ 3.
a Find the gradients at the beginning and the end of the trail.
b Calculate the point where the gradient is zero, and calculate also the height of the pass.
5 A tadpole begins to swim vertically upwards in a pond and after t seconds it is
25 − 0.1t3 cm below the surface.
a How long does the tadpole take to reach the surface, and what is its speed then?
b What is the average speed over this time?
6 a Show that the gradients of the curve y = x(x − 2) at the points (0, 0) and (2, 0) only
differ in sign. What is the geometrical interpretation for this?
b If the gradient of the curve y = x(x − 2)(x − 5) at the points (0, 0), (2, 0) and (5, 0) are
1
1
1
l, m and n respectively, show that + + = 0.
l
m
n
Cambridge University Press • Uncorrected Sample Pages •
2008 © Evans, Lipson, Wallace TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard