13
syllabus
syllabus
rref
efer
erence
ence
Topic:
• Rates of change
• Optimisation using
derivatives
In this
In this
cha
chapter
pter
13A Introduction to limits
13B Limits of discontinuous,
rational and hybrid
functions
13C Differentiation using first
principles
13D Finding derivatives by rule
13E Rates of change
13F Solving maximum and
minimum problems
Differentiation
and
Introduction
In the previous chapter we considered the problem of determining the speed of a spanner as it falls from a high tower. Such a calculation is not straightforward because the speed of the spanner is constantly changing. We modelled the situation by drawing a graph of the distance, d, the spanner had travelled versus time, t.
We noted that the instantaneous speed of the spanner at t = 3 seconds was the gradient of the tangent to the curve at t = 3 seconds.
In this case, the gradient of the tangent is:
= 30 m/s
That is, the instantaneous speed of the spanner at t = 3 seconds is 30 m/s.
In this chapter we will extend this concept of finding the instantaneous rate of change. Suppose we know that the equation relating the distance, d, the spanner falls with time, t, is:
d = 5t2
Taking the techniques developed previously we can now give an expression for the gradient of the tangent at any point on the curve between t = 0 and t = 3.
What is the gradient at the point, P; that is, at t seconds?
Consider the chord joining P(t, 5t2) and Q(t + 0.01, 5(t + 0.01)2). 45–0
3–1.5
---t (seconds)
d
(metres)
0 10 20 30 40 50
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
t (seconds) P (t,5t2)
Q (t+0.01,5(t+0.01)2)
d
(metres)
0 10 20 30 40 50
The gradient of this chord is:
Gradient = 5(t + 0.01)2− 5t2 = 5t2 + 0.1t + 0.0001 − 5t2 = 10t + 0.01
We can see that as Q approaches P the speed tends to 10t exactly. The function 10t is called the gradient function of 5t2.
The limit
Look carefully at the words used in the previous solution: ‘As Q approaches P the speed tends to 10t exactly’.
The point Q never quite reaches P and the speed never quite reaches 10t. However, they do get arbitrarily close — as close as you can get.
This is the limit concept in informal terms.
Expressing limits in mathematical language, we say that a limit can be used to describe the behaviour of a function, f(x), as the independent variable, x, approaches a certain value, say a. In some cases the function will not be defined at a. Using the cor-rect notation for the earlier example on the gradient of the chord, PQ, we write:
This is read as ‘the limit of f(x) as x approaches t is equal to 10t’.
Limits of continuous functions
A continuous function has a graph that forms a continuous line; that is, it has no breaks. If a function is continuous at the point where a limit is being found then the limit always exists and can easily be found by direct substitution.
Add the following series of numbers and state what value it is approaching.
+ + + + + . . .
THINK WRITE
Add the first 2 terms. Sum of first 2 terms is (= 0.750).
Add the first 3 terms. Sum of first 3 terms is (= 0.875).
Add the first 4 terms. Sum of first 4 terms is (≈ 0.938).
Add the first 5 terms. Sum of first 5 terms is (≈ 0.969).
Add the first 6 terms. Sum of first 6 terms is (≈ 0.984).
Give the upper limit. The sum is approaching 1.
1 2 --- 1
4 --- 1
8 --- 1
16 --- 1
32
---1 34
---2 78
---3 1516
---4 3132
---5 6364
---6
1
WORKED
Example
f x( ) x→t
lim = 10t
By investigating the behaviour of the function f(x) = x + 3 in the vicinity of x = 2 show that .
THINK WRITE
Create a table of values for x and f(x) in the vicinity of x = 2.
Consider the values taken by f(x) as x approaches 2.
As x approaches 2 from the left and the right, f(x) approaches a value of 5. So
f x( )
xlimÆ2 = 5
1
x 1.95 1.99 1.995 2 2.005 2.01 2.05
f(x) 4.95 4.99 4.995 5 5.005 5.01 5.05
2 y
x 0 3 5
–3 2
f(x)
f x( )
xlim→2 = 5.
2
In general if a function, f(x), is continuous when x=a, then .
Theorems on limits
Theorem 1
For the constant function f where f(x) = c, then .
If and , then
Theorem 2
= = A ± B
For example: =
= 1 + 2 = 3
Theorem 3
= = A × B
For example: =
= 1 × −3 = −3
Theorem 4
= if B ≠ 0
For example: =
=
These theorems on limits can be expressed in word form:
Find .
THINK WRITE
To consider whether the function is continuous, sketch the graph of y = x2+ 2 in the vicinity of x = 2.
As the graph is continuous, substitute x = 2 into x2+ 2 to evaluate the limit.
The function is continuous at x = 2.
= 22+ 2
= 6 x2+2
( )
xÆ2
lim
1 y
x 0 2 6
2 f(x)
2
x2+2
( )
xlim→2
3
WORKED
Example
f x( )
xlimÆa = f a( )
f x( ) xlim→a = c f x( )
x→a
lim = A g x( ) x→a
lim = B
f x( )±g x( )
[ ]
x→a
lim f x( )
x→a
lim g x( ) x→a
lim ±
x2+2x
( )
xlim→1 x
2
xlim→1 +xlim→12x
f x( ) × g x( )
[ ]
x→a
lim f x( )
x→a
lim × g x( ) x→a
lim x2( )3x
[ ]
x→–1
lim ( )x2
x→–1
lim × ( )3x x→–1
lim
f x( ) g x( ) ---x→a
lim
f x( ) xlim→a
g x( ) x→a
lim
--- A B ---=
x2+4 x+1 ---xlim→5
x2+4
( )
x→5 lim
x+1
( )
xlim→5 ---29
---1. The limit of a sum = the sum of the limits.
2. The limit of a difference = the difference of the limits. 3. The limit of a product = the product of the limits. 4. The limit of a quotient = the quotient of the limits.
Introduction to limits
1 Add the following series of numbers and state what value it is approaching. 4 + 2 + 1 + + + . . .
2 The diagram at right shows regular polygons with 3, 4 and 5 sides. As the number of sides gets very large (→ ∞) what shape emerges?
3
If n represents the number of sides of a regular polygon, then which of the following correctly describes the situation given in question 2 above?
4 a Find the value of as n gets infinitely large.
b Write this using limit notation.
5 a If S = 1 + + + + + . . . and n represents the number of terms to be summed in the series, copy and complete the following table:
b
Which of the following is equal to ?
A n B C D n → 0 E n → 5
n 1 2 3 4 5 6 10
S 1 1
A 1.75 B 1.95 C 2 D 1 E 0
remember
1. If a function, f(x), is continuous when x=a, then .
2. For the constant function f where f(x) =c, .
3. =
4. =
5. =
f x( )
x→a
lim = f a( )
f x( )
x→a
lim = c
f x( )±g x( )
[ ]
x→a
lim f x( )
x→a
lim g x( )
x→a
lim ±
f x( ) × g x( )
[ ]
xlim→a xlim→af x( ) × xlim→ag x( )
f x( ) g x( )
---x→a
lim
f x( )
x→a
lim
g x( )
x→a
lim
---remember
13A
W WORKEDORKED
E Example
1 12 ---1 4
---n = 3 n = 4 n = 5 m
multiple choiceultiple choice
n→∞ lim
n→0 lim 1
n
---1 2 --- 1
4 --- 1
8 --- 1
16
---1 2−
m
multiple choiceultiple choice
6 By investigating the behaviour of the function f(x) =x + 5 in the vicinity of x = 3
show that .
7
From the graph at right it can be seen that is equal to:
8 For each of the functions graphed below find .
9 Evaluate the limits below.
10
If f(x) = 3x − 2 then the value of is equal to:
11
The is equal to:
12 Find the value of the following limits.
A 2 B –1 C 3 D 1 E 0
a b c
d e f
a b c
d e f
g h i
j k l
A 0 B 3 C 2 D −2 E 1
A 4 B −5 C 9 D –4 E 0
a b c d
W WORKEDORKED
E Example
2 x→3f x( )
lim = 8
SkillS
HEET
13.1
y x 0 3 2 1 –1 2
f (x)
m
multiple choiceultiple choice
f x( ) x→2
lim
f x( ) x→4
lim y x 0 30 14 4
f (x) y
x 0
–3
4
f(x)
y
x 0
2
4 f(x)
y x 0 21 5 4 f(x)
y x 0 4 1 4
f (x)
y x 0 10 15 4
f (x)
W WORKEDORKED
E Example
3
Math cad
Limits
2x+5
( )
x→2
lim (10–2k)
k→4
lim (7p–4)
p→–1 lim 49a+1
( )
a→1
lim (x2+2)
xlim→0 x
2–7
( )
xlim→2 h2+3h
( )
h→5
lim (8–5x2)
x→2
lim (h2+h–4)
h→1 lim 3p2–2p+3
( )
plim→–2 x
3+2x
( )
xlim→0 t
3+5t–9
( )
tlim→–2 m
multiple choiceultiple choice
f x( ) xlim→0
m
multiple choiceultiple choice
x2–5
( )
x→–3 lim
x2+3 x ---x→2
lim 2x
2–3x x+1 ---x→0
lim 6x–2
x ---x→–2
lim x2+3x–2
x–2 ---x→3
Limits of discontinuous, rational and
hybrid functions
Limits of discontinuous functions
If a function is discontinuous at the point where the limit is being investigated then the limit will exist only if the function is approaching the same value from the left as from the right. Consider the discontinuous functions graphed below.
1. From the left, .
(The symbol x → 1− indicates that we are letting x approach 1 from the left side.)
From the right, . Left limit = right limit.
Therefore, . 2. From the left, .
From the right, . Left limit ≠ right limit.
Therefore, does not exist.
Sneaking up on a limit
1 Consider the expression .
a Complete a table of values by evaluating the expression for x = 1.5, 1.2, 1.1, 1.01, 1.001, 1.0001 and 1.
b Copy and complete:
As x gets closer to 1 from above, gets closer to , or
= .
c Complete a table of values by evaluating the expression for x = 0.5, 0.6, 0.9, 0.95, 0.99, 0.999 and 1.
d Copy and complete:
As x gets closer to 1 from below, gets closer to , or
= .
e Explain why the limit can be evaluated, but not the actual value of the expression above when x = 1.
2 Use a similar technique to evaluate the limit of as x approaches 3.
3 Find .
x2+x–2 x–1
---x2+x–2 x–1 ---x2+x–2
x–1 ---xlim→1+
x2+x–2 x–1 ---x2+x–2
x–1 ---xlim→1–
x2+2x–15 x–3 ---x2–7x
x–7 ---x→7
lim
y
x 0
1
1 f x( )
x→1−
lim = 1
f x( ) xlim→1+ = 1 f x( )
x→1
lim = 1
y
x 0
2 4
1 2 f(x) f x( )
x→1−
lim = 2
f x( ) x→1+
lim = 4
Limits of rational functions
Finding the limit of a rational function involves simplifying the function before direct substitution can take place.
a Find .
b Sketch the graph of , stating the value of x for which it is discontinuous.
THINK WRITE
a We cannot substitute x = 1 directly as we will get which is undefined, so we factorise the numerator.
a =
Cancel out the common factor. = , x ≠ 1
Now substitute x = 1 and evaluate. = 1
b The graph of is the same as the graph of f(x) = x, except where the point (1, 1) does not exist.
b
The function is discontinuous at x = 1. x2–x
x–1
---xÆ1
lim
f x( ) x
2–x
x–1
---=
1
0 0
---x2–x
x–1
---x→1
lim x x( –1)
x–1
---x→1
lim
2 x
x→1
lim
3
f x( ) x 2–x x–1
---= y
x 0
1
1 f(x)
4
WORKED
Example
a By first factorising the numerator, simplify the rational function ,
stating the value for which the function does not exist (that is, is discontinuous).
b Find , where a is the value at which f(x) is discontinuous.
THINK WRITE
a Factorise the numerator. a
Cancel out the common factor. , x ≠ −3
so f(x) is discontinuous at x = –3
b Write an expression for and evaluate by substituting x = −3.
b = −3 + 2
= −1
f x( ) x2+5x+6 x+3
---=
f x( )
xlimÆa
1 f x( ) x
2+5x+6
x+3
---=
x+3
( )(x+2) x+3
( )
---=
2 f x( ) = x+2
f x( )
xlim→a xlim→–3(x+2)
5
Limits of hybrid functions
Hybrid functions are functions that have different rules for different parts of the domain.
Dirichlet’s function
The French mathematician Dirichlet devised the following hybrid function. D(x) =
So D(0.5) = 0 but D( ) = 1
Can you decide the value of ?
a Sketch the graph of the hybrid function f(x) = x2, x ∈ (−∞, 2] x − 1, x ∈ (2, ∞)
b Find i ii iii if it exists.
THINK WRITE
a Sketch f(x) = x2 over the domain (−∞, 2].
a
Sketch (on the same axes)
f(x) = x − 1 over the domain (2, ∞).
b i Substitute x = 2 into f(x) = x2. b i = = 4 ii Substitute x = 2 into f(x) = x − 1. ii =
= 1
iii Are these limits equal? iii does not exist (as left limit
≠ right limit). f x( )
xlimÆ2− xlimÆ2+f x( ) xlimÆ2f x( )
1 y
x 0
1 4
2 f(x) 2
f x( )
x→2−
lim x2
x→2−
lim
f x( )
x→2+
lim (x–1)
x→2+
lim
f x( )
x→2
lim
6
WORKED
Example
{
0 if x is a rational number 1 if x is not a rational number
2
D x( ) x→2
lim
remember
1. If a function is discontinuous at the point where the limit is being investigated then the limit will exist only if the function is approaching the same value from the left as from the right.
2. Finding the limit of a rational function involves simplifying the function before direct substitution can take place.
3. Hybrid functions are functions that have different rules for different parts of the domain.
Limits of discontinuous,
rational and hybrid functions
1 Which of the following graphs are discontinuous?
2 For each discontinuous function above state the value of x for which it is discontinuous.
3 a Find .
b Sketch the graph of , stating the value of x for which it is discontinuous.
4 a Evaluate f(x) when x= 0 if . Comment on this result.
b For what value of x is f(x) discontinuous?
c Factorise the numerator of f(x).
d Now simplify f(x).
e Sketch the graph of f(x).
f Evaluate if it exists.
5 By first factorising the numerator, simplify the following rational functions, stating the value for which the function does not exist (is discontinuous).
a b c
d e f
a b
c d
e f
g h
13B
y
x 0
3
3
y
x
0 4
y
x 0
y
x 0
–1
y
x 0
4
2
y
x
0 3
W WORKEDORKED
E Example
4
x2–4x x–4 ---x→4
lim
f x( ) x 2–4x x–4 ---=
f x( ) x2+x x ---=
f x( ) x→0
lim
W WORKEDORKED
E Example
5a
f x( ) x 2+3x
x
---= f x( ) 6x–18
x–3 ---=
f x( ) x 2–5x
x
---= f x( ) x
2+5x+4 x+4 ---=
f x( ) x2–7x+6 x–6
---= f x( ) x3+8
x+2 ---=
f x( ) x2+3x–4 x–1
---= f x( ) x3–27
6 For each rational function in question 5 above find , where a is the value at which f(x) is discontinuous.
7 Sketch the graphs of the following hybrid functions.
8 For each of the corresponding functions in question 7 above evaluate the following.
9 Investigate whether the following limits exist. For those that do exist state the limit.
10 Evaluate the following.
a b
c d
a i ii iii
b i ii iii
c i ii iii
d i ii iii
a b c
d e f
a b c d e f g h i j k l W WORKEDORKED E Example 5b
f x( ) x→a
lim W WORKEDORKED E Example 6a
f x( ) x+3,x∈(–∞,2) 4,x∈[2,∞)
= g x( ) 4–x x, ∈(–∞,–1] 3x+1,x∈(–1,∞)
=
h x( ) 2x x, ∈(–∞,0) x2,x∈[0,∞)
= p x( ) x
2
1,x∈(–∞,1] +
x+2,x∈(1,∞) = W WORKEDORKED E Example
6b f x( )
xlim→2− xlim→2+f x( ) xlim→2f x( )
g x( ) x→–1−
lim g x( )
x→–1+
lim g x( )
x→–1 lim h x( )
x→0−
lim h x( )
x→0+
lim h x( )
x→0 lim p x( )
x→1−
lim p x( )
x→1+
lim p x( )
x→1 lim
x+2,x∈(–∞,2) 3,x∈[2,∞)
x→2
lim x–4,x∈(–∞,0]
x–4,x∈(0,∞)
x→2
lim 5–x x, ∈(–∞,3)
2x+1,x∈[3,∞)
x→3
lim
2x–3,x∈(–∞,1] 2x+1,x∈(1,∞)
x→1
lim x
2
x∈(–∞,2] ,
x+2,x∈(2,∞)
x→1
lim 4 x
2
x∈(–∞,–2) ,
–
x+2,x∈[ 2– ,∞)
x→–2
lim
x2+3x+2
( )
x→2
lim 14x+7
2x+1 ---x 1 2 ---– → lim
3x+1
xlim→1 ,x∈(–∞,1) x2+3,x∈[1,∞ )
{
x→–1---x2–x9x+1–10 lim12–3x–2x2
( )
xlim→2
x2–2x+1,x∈(–∞,1] x+3,x∈(0,∞)
xlim→0 x2+3x+2
x+2 ---xlim→–1
x3+x2–5x
( )
x→1 lim x+3
( )(x–3) x–3 ---xlim→3
x+2
( )(2x–3) x+2 ---xlim→–2
x2+7x+6 x+1 ---x→–1
lim x2–6x+8
x–4 ---x→4
Differentiation using first principles
The gradient function is the rule for the instantaneous rate of change of a given func-tion at any point. The gradient at any point (x, y) can be found by substitufunc-tion into the gradient function.
Consider the secant PQ drawn to the curve f(x) below.
The coordinates of P are [x, f(x)] and the coordinates of Q are [x + h, f(x + h)]. So the gradient of the secant PQ =
=
=
As h approaches zero, that is, as Q draws as close as possible to P along the curve, PQ effectively becomes a tangent to the curve at P.
We can therefore say that the gradient of the tangent at P is or , h ≠ 0,
where f′(x) denotes the gradient of a tangent at any point, x, on the graph of f(x). That is, f′(x) is the gradient function of f(x).
The process of finding the gradient function is called differentiation from first principles, or finding the derivative from first principles.
Two different forms of notation are commonly used to represent a function and its derivative.
1. The European notation of Leibniz is: 2. The alternative notation is: (a) y for the function (a) f(x) for the function
(b) for the derivative. (b) f′(x) or for the derivative. y
x 0 x x + h f(x + h)
f(x)
y = f(x) Q
P
Tangent at P
rise run
---f x( +h)– f x( ) x+h–x
---f x( +h)– f x( ) h
---y
x
0 h
Q
Q
Q P
Q moves closer to P as h approaches 0
f x( +h) – f x( ) h ---h→0
lim
f′( )x f x( +h)– f x( ) h ---hlim→0
=
f x( +h)– f x( ) h ---h→0
lim
dy dx
--- d
Secants and tangents
A spreadsheet may be used to study the effect h has on the gradient of a secant, by calculating f(x + h), f(x) and h, as well as the gradient .
Use a spreadsheet (you may wish to use the Maths Quest spreadsheet ‘Gradient of a secant’) to answer the following.
1 For f(x) = x2+ 1, find the gradient of the secant between:
a x = 1 and x = 2
b x = 1 and x = 1.5
c x = 1 and x = 1.1
d x = 1 and x = 1.01
e x = 1 and x = 1.001.
2 Use your answers to question 1 to predict the gradient of the tangent to f(x) = x2+ 1 at x = 1.
3 For f(x) = –x2+ 4x (enter this as –(x2) + 4*x if using the Maths Quest spreadsheet), find the gradient of the secant between:
a x = 0 and x = 2
b x = 0 and x = 1
c x = 0 and x = 0.5
d x = 0 and x = 0.1
e x = 0 and x = 0.001.
4 What is the gradient of the tangent to f(x) = –x2+ 4x at x = 0?
E
XCEL
Spreadshe et
Gradient of a secant
Mathc
ad
Gradient of a secant
f x( +h) – f x( ) h
---Find the derivative of x2− 2x using first principles.
THINK WRITE
Define f(x). f(x) = x2− 2x
The derivative is equal to: .
f′(x) =
Simplify the numerator f(x + h) − f(x). f(x + h) − f(x)
= (x + h)2− 2(x + h) − (x2− 2x)
= x2+ 2xh + h2− 2x − 2h − x2+ 2x
= 2xh + h2− 2h
Factorise the numerator f(x + h) − f(x). = h(2x + h − 2)
Simplify by
cancelling the common factor of h.
=
=
Evaluate the limit by substituting h = 0. = 2x − 2.
1
2
f x( +h)– f x( ) h ---h→0
lim
f x( +h)– f x( ) h
---h→0
lim
3
4
5 f x( +h)– f x( ) h ---hlim→0
f x( +h)– f x( ) h
---h→0
lim
h(2x+h–2) h
---h→0
lim
2x+h–2
( )
h→0
lim ,h≠0
6
7
Note: For all polynomial functions, f(x): when the expression f(x + h) − f(x) is simpli-fied all of its terms have h as a factor.
If g(x) = 2x2+ 5x − 2, find:
a g′(x) using first principles b the value(s) of x where the gradient equals 0.
THINK WRITE
a Let g(x) = 2x2+ 5x − 2. a g(x) = 2x2+ 5x − 2 The derivative is equal to: g′(x) =
Simplify the numerator g(x + h) − g(x).
g(x + h) − g(x)
= 2(x + h)2+ 5(x + h) − 2 − (2x2+ 5x − 2)
= 2(x2+ 2xh + h2) + 5x + 5h − 2 − 2x2− 5x + 2
= 2x2+ 4xh + 2h2+ 5x + 5h − 2 − 2x2− 5x + 2
= 4xh + 2h2+ 5h
Factorise the numerator g(x + h) − g(x).
= h(4x + 2h + 5)
Simplify by
cancelling the common factor of h.
=
=
Evaluate the limit by substituting h = 0.
= 4x + 5 So g′(x) = 4x + 5.
b Solve g′(x) = 0. b g′(x) = 0 4x + 5 = 0 4x = −5
x = −
So the gradient equals 0 when x = − .
1
2
g x( +h) g x– ( ) h ---h→0
lim
g x( +h) – g x( ) h
---hlim→0
3
4
5 g x( +h) – g x( ) h ---h→0
lim g x( +h)–g x( )
h
---h→0
lim
h(4x+2h+5) h
---h→0
lim
4x+2h+5
( )
h→0
lim ,h≠0
6
5 4
---5 4
---8
WORKED
Example
remember
1. The process of finding the gradient function is called
differentiation from first principles.
2. Differentiating y gives .
3. Differentiating f(x) gives f′(x).
f x( +h)– f x( ) h
---h→0 lim
dy
dx
Differentiation using first
principles
1 Find the derivative of the following using first principles.
2 Use first principles to find .
3 If g(x) = x2− 6x find:
a g′(x) using first principles
b the value(s) of x where the gradient equals 0.
4 a If f(x) = x3− 8, find f′(x) using first principles.
b Hence, determine the value(s) of x where the gradient function is equal to 12.
5 By first deriving the gradient function f′(x), evaluate f′(3) when f(x) is equal to:
6
Which of the following do not denote the gradient at any point on a graph? (One or more answers.)
7
The most accurate method for finding the gradient when x = 3 for the function f(x) = x2+ 2x is by:
A sketching the graph and drawing a tangent at x = 3 to find the gradient
B finding the gradient of the secant to the curve joining the points where x = 3 and x = 3.1
C finding f′(x) using first principles and evaluating f′(3)
D guessing
E finding the gradient of the line from the origin to the point (3, 15).
8
Given that f′(x) = 4x if f(x) = 2x2 and g′(x) = 3x2+ 1 if g(x) = x3+ x, then the deriv-ative of x3+ 2x2+ x must be equal to:
a 5x − 7 b x2+ 10x c x2− 8x d x3+ 2x
a y = x2+ 3 b y = x2 − 3x + 1 c y = 4x2
d y = 9 − x2 e y = 6x − 2x2 f y = x3+ 5x − 4
a 7x + 5 b x2+ 4x c x2− 3x + 2 d x3− 5.
A f′(x) B C
D E
A 3x2+ 4x + 1 B 12x3+ 4x C 3x2+ 4x
D 2x5+ 2x3 E 5x2+ 1
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f x( +h)– f x( ) h ---hlim→0
f x( +h)– f x( ) h ---hlim→∞
dy dx
--- f x( +h)– f x( ) h
---m
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Work
Finding derivatives by rule
Fortunately the tedious process of finding derivatives from first principles need not be applied once rules are established. For polynomial functions the following rules apply.
Rule 1. If f(x) = xn, then f′(x) = nxn − 1.
Rule 2. If f(x) = axn, then f′(x) = naxn − 1.
Rule 3. If f(x) = c, then f′(x) = 0 (where c is constant).
Rule 4. If f(x) = g(x) + h(x), then f′(x) = g′(x) + h′(x).
Differentiate each of the following:
a b c d .
THINK WRITE
a Write the expression for y. Apply rule 1 to find the derivative.
a y = x8
b Apply rule 2. b y = 3x2
c Apply rules 2, 3 and 4. Remember that x0= 1.
c y = 7x + 3
d Differentiate the 3 terms separately (that is, apply rules 2 and 4).
d y = 2x5 + x2 – 6x
y = x8 y = 3x2 y = 7x+3 y 2x5 3
5 ---x2–6x
+ =
dy dx
--- = 8x8–1
8x7
=
dy dx
--- = 2 3( x2–1)
6x
=
dy dx
--- = 7x1–1+0
7x0
=
7
=
3 5
---dy dx
--- 5 2( x5–1) 2 3 5 ---x2–1
( )–1 6x( 1–1)
+ =
10x4 6
5 ---x–6x0
+ =
10x4 6
5 ---x–6
+ =
9
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Example
Find f′(x) if f(x) = 3x(x − 2).
THINK WRITE
Write down f(x). f(x) = 3x(x − 2)
Expand the brackets. f(x) = 3x2− 6x Differentiate by rule. f′(x) = 6x − 6
1 2
3
10
If , find g′(x).
THINK WRITE
Factorise the numerator because at this stage we can only differentiate a constant denominator.
Simplify g(x). = x(4x + 3), x ≠ 0
Expand the brackets. = 4x2+ 3x
Differentiate g(x) by rule. g′(x) = 8x + 3 g x( ) 4x
3+3x2
x
---=
1 g x( ) 4x
3+3x2
x
---=
x2(4x+3)
x ---= 2 3 4
11
WORKED
Example
Differentiate each of the following:
a b c d .
THINK WRITE
a Write down f(x). Differentiate by rule 1.
a f(x) = x–3 f′(x)= −3x−3 − 1
= −3x−4 b Write down f(x).
Bring the x term to the numerator using the index laws, as we can only differentiate a constant denominator.
b f(x)=
= 1x−7
Differentiate by rule 1. f′(x)= −7(1x−7 − 1)
= −7x−8
Express answer with a positive index to follow style of f(x). =
c Write down f(x). Differentiate by rule 1.
c f(x) = f′(x)=
=
Express answer with a positive index. =
d Write down f(x).
Convert x to index form.
d
f(x) =
Bring the x term to the numerator using the index laws. f(x) =
Differentiate by rule 2. f′(x)= −
(
)
=
Express with a positive index. =
Express the power of x back in surd (square root) form. =
f x( ) = x–3 f x( ) 1
x7
---= f x( ) x
1 3
---= f x( ) 4
x ---= 1 2 1 2 1 x7 ---3 4 7 x8 ---– 1 2
x13
---1 3 ---(x
1 3
---–1
) x– 2 3 ---3 ---3 1 3x 2 3 --- ---1 2
f x( ) 4 x ---= 4 x 1 2 --- ---3 4x 1 2 ---– 4 1 2 --- 4x 1 2
---– –1
Tangents and normals
As mentioned earlier in this chapter the derivative f′(x) is actually the gradient function. This means that the value of the gradient at any particular point on a curve is equal to the numerical value of the derivative at that point.
Recall that if the gradient of a tangent to a curve at point P is mT, then the normal, mN, is a straight line perpendicular (at right angles) to the tangent such that mN= − and passing through the point P as shown at right.
Also recall that the equation of a straight line is given by y − y1= m(x −x1) where (x1, y1) is the point P, above, and m is the gradient.
y
x 0
P
Tangent Normal
1 mT
---a Find the equation of the tangent to the curve f(x) = x2+ 6x − 8 at the point where the gradient has a value of 8.
b Hence, find the equation of the normal at this point.
THINK WRITE
a Find the gradient function of the curve, f′(x).
a f(x) = x2+ 6x – 8 f′(x) = 2x + 6
Find x1, the value of x where f′(x) = 8; that is, solve 2x + 6 = 8.
For gradient = 8 2x + 6 = 8 2x = 2
x = 1 So x1= 1.
Find f(x1) to determine the value of y1.
y1= f(x1)
= f(1)
= (1)2+ 6(1) − 8
= −1
Simplify the equation
y − y1= mT (x − x1) to find the equation of the tangent.
The equation of the tangent at the point (1, −1) is y − −1 = 8(x − 1)
y + 1 = 8x − 8 y = 8x − 9 b Find the gradient of the normal using
.
b mN= −
Simplify the equation
y − y1= mN (x − x1) to find the equation of the normal.
The equation of the normal at the point (1, −1) is y − −1 = − (x − 1)
y + 1 = − 8y + 8 = −x + 1 x + 8y + 7 = 0
1
2
3
4
1
mN 1
mT ---– =
1 8
---2
1 8
---x–1
( )
8
---13
Finding derivatives by rule
1 Differentiate each of the following.
2 Find if y is:
3 Match the correct derivative from the set A to G below to each of the following.
4 Differentiate the following.
5 Find f′(x) if f(x) is:
6 Find f′(x) for each of the following.
7 Find g′(x) by first simplifying g(x).
a y = x6 b y = 7x2 c y = 5x + 2 d y = 4x4+ x2− 5x
a 3x4 b 8x7 c 5x5 d −4x6 e −2x3 f −7x
a x8 b x4+ x2 c 2x3− 4x + 7 d x2+ 6x − 5
e x4+ 2x3− 3 f x5+ 6x3− 4x g xp
A 5x4+ 18x2− 4 B 4x3+ 2x C pxp − 1 D 8x7
E 6x2− 4 F 2x + 6 G 4x3+ 6x2
a y = x6+ 3x2− 4 b y = 5x4− 7x3+ 6x
c y = x11− 3x6+ 4x5+ 3x2 d y = 10x5 − 3x4+ 2x3− 8x
e y = 6 f y = 3x4+ 5x4
a x4 b − x2 c x7
d x3+ x2− 3x e x5+ x4+ x3 f 4x3− x2− x + 8
a f(x) = x(x + 3) b f(x) = 3x(2x − 5) c f(x) = (x + 4)2
d f(x) = 9(8 − 3x)2 e f(x) = (x + 2)3 f f(x) = (2x − 5)3
a b
c d
remember
1. Rule 1: If f(x) =xn, then f′(x) =nxn− 1. 2. Rule 2: If f(x) =axn, then f′(x) =naxn− 1.
3. Rule 3: If f(x) =c, then f′(x) = 0 (where c is constant).
4. Rule 4: If f(x) =g(x) +h(x), then f′(x) =g′(x) +h′(x).
5. Equation of tangent: y−y1=mT (x−x1)
6. Equation of normal:y−y1=mN (x−x1)
where mN= − 1
mT
---remember
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Derivatives
2 3
---dy dx
---2 3
--- 5
8
--- 6
7
---3 4 --- 1
2
--- 2
5 --- 3
4 --- 1
6
--- 4
7 --- 5
6
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11 g x( ) x3+5x
x
---= g x( ) 8x2–6x
2x ---=
g x( ) 3x
3+2x2–5x x
---= g x( ) 5x
8 Differentiate each of the following:
9 Evaluate i f′(1) ii f′(−2) and iii f′(0) for each of the following.
10 a Find the x-intercepts of the parabola y = x2− 5x + 6.
b Find the gradient of the parabola at the points where it crosses the x-axis.
c Determine the value of x for which the gradient of the parabola is:
i 0
ii 7
iii −3.
11 a Find the x-intercepts of the curve y = 2x2+ 5x − 3.
b Find the gradient of the curve at these points.
c Find the coordinates of the point where the gradient is 0.
12 Find the coordinate(s) of the points on the curve x3− 3x2 where the tangent:
a is parallel to the x-axis
b is parallel to the line y = −3x + 2.
13 a Find the equation of the tangent at the point on the curve x2+ 4x − 1, where the gradient is 6.
b Hence, find the equation of the normal at this point.
14 Find the equation of the normal to the curve, y = 2x2− 2x + 5, at the point where the curve crosses the y-axis.
15 Find the equation of the normal to the curve y = –x2+ 4x at:
a x = 2
b x = 1.
16 Find the equation of the normal to the graph of y = x3+ 2x2 – 3x + 1 at x = –2.
a x−4 b x−7 c 3x−4
d 5x−8 e −4x−6 f −3x−5
g h i
j k l
m n o
p q r
s
a f(x) = 5x2+ 3x − 1
b f(x) = x3+ 2x2− 4
c f(x) = 3x2− 2x + 6
d f(x) = x3+ 7x − 8
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1 x4
--- 1
x9
--- 5
x3 ---10
x6
--- 2x12--- x
2 3
---4x
1 4
---3x
2 5
---x
1 x
--- 4 x 3 x
2 x 3
---1 3
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Gradient at a point
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Gradient at a point
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Tangent and normal
EXCEL Spreadshe
et
1. Enter a function for Y1 in the Y= menu.
2. Move to the Y2 position and press
then select 8:nDeriv(.
3. Fill in the arguments for nDeriv as follows:
Y1, X, X.
(Y1 is found under VARS/Y-VARS/Function.)
4. Set suitable WINDOW values and press to view the function (Y1) and its
derivative (Y2). Press to study values,
using the up arrow to swap between graphs.
Graphs of derivatives
Now that we have a quick way to find the rule for the derivative f′(x) or of a function f(x), we may graph the derivative using its rule, rather than by finding several gradients and plotting points.
For example, if f(x) =x3+ 12x, then
f′(x) = 3x2 – 12 = 3(x2 – 4) = 3(x+ 2)(x – 2).
f(x) and f′(x) may be plotted on the same axes as shown at right.
1 Use technology to produce graphs of the following functions and their derivatives, with each pair on the same set of axes. Show all turning points of the original function.
a f(x) =x3 – 5x2+ 1
b f(x) = 2x2+ 4x + 5
c f(x) =x4 – 3x3– 3x2+ 11x – 6
d f(x) = 6x – 3
2 Copy and complete: The of the original function (thick graph) are above or below the x of the derivative graph.
x 0
40
20
–40 –20
5 –5
y f'(x)
f(x)
dy
dx
---EXCE L Spreadsheet
EXCE L Spreadsheet
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Graphics Calculator
Graphics Calculator
tip!
tip!
Plotting the derivative function
CASIO
Derivative
function MATH
GRAPH
TRACE
Rates of change
The rate of change of a function refers to the rate at which its gradient is changing. For linear functions the gradient is constant; however, the gradient for other functions such as quadratic or cubic polynomials is continually changing.
Differentiation provides us with a tool to describe the gradient of a function and hence determine its rate of change at any particular point. In essence, while average rates of change can be determined from the original function, differentiation of this function provides a new function that describes the instantaneous rate of change. Note: The term instantaneous rate of change is often referred to as rate of change.
The rate of change of velocity with respect to time is acceleration.
➙
If f(x) = x2− 2x + 4 find:
a the average rate of change between x = 2 and x = 4
b a new function that describes the rate of change
c the instantaneous rate of change when x = 4.
THINK WRITE
a Write the function. a f(x) = x2− 2x + 4
Average rate of change = . Average rate of change=
= = 4
b Differentiate f(x). b f′(x) = 2x − 2 c Substitute x = 4 into f′(x). c f′(4)= 2(4) − 2
= 6
So the rate of change when x = 4 is 6.
1
2 change in f x( )
change in x
--- f( )4 – f( )2
4–2
---12–4 2
---14
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Example
The rate of change of position with respect to
It is worth noting that there are two common ways of writing the derivative as a func-tion. For example, the derivative of the function P(x) = x2+ 5x − 7 may be written as P′(x) = 2x + 5 or as .
A javelin is thrown so that its height, h metres above the ground, is given by the rule:
h(t) = 20t − 5t2+ 2, where t represents time in seconds.
a Find the rate of change of the height at any time, t.
b Find the rate of change of the height when i t = 1 ii t = 2 iii t = 3.
c Briefly explain why the rate of change is initially positive, then zero, and then negative over the first 3 seconds.
d Find the rate of change of the height when the javelin first reaches a height of 17 metres.
THINK WRITE
a Write the rule. a h(t) = 20t − 5t2+ 2
Differentiate h(t). h′(t) = 20 − 10t b Evaluate h′(1). b h′(1)= 20 − 10(1)
= 10 m/s
Evaluate h′(2). h′(2)= 20 − 10(2)
= 0 m/s
Evaluate h′(3). h′(3)= 20 − 10(3)
= −10 m/s
c For rates of change: Positive means increasing.
Zero means neither increasing nor decreasing. Negative means decreasing.
c The javelin travels upward during the first 2 seconds.
When t = 2 seconds the javelin has reached its maximum height. When t > 2 seconds the javelin is travelling downward.
d Find the time at which a height of 17 m occurs, by substituting h = 17 into h(t).
d 20t − 5t2+ 2 = 17
Make RHS = 0. −5t2 + 20t − 15 = 0
Solve for t.
Note: the quadratic formula could also be used to solve for t.
t2− 4t + 3 = 0 (t − 1)(t − 3) = 0
t = 1 or 3
The first time it reaches 17 m is the smaller value of t.
It first reaches 17 m when t = 1 s.
Evaluate h′(1). h′(1)= 20 − 5(1)2
= 15 m/s
Rate of change of height is 15 m/s.
1 2
1
2
3
1
2
3
4
5
15
WORKED
Example
dP dx
The shockwave from a nuclear blast spreads out at ground level in a circular manner.
a Write down a relationship between the area of ground, A km2, over which the shockwave passes and its radius, r km.
b Find the rate of change of A with respect to r.
c Find the rate of change of A when the radius is 2 km. d What is the rate of change
of A when the area covered is 314 km2?
THINK WRITE
a State the formula for the area of a circle. a A(r) =πr2
b Differentiate A(r). b A′(r) = 2πr
c Substitute r= 2 into A′(r).
Note: The units for the rate of change of
A (km2) with respect to r (km) are
km2 per km or km2/km.
c A′(2)= 2(3.14)(2)
= 12.56
Rate of change of A when the radius is 2 km is 12.56 km2/km.
d Substitute A= 314 into the area
function A(r) and solve for r.
d A(r) =πr2
314 = 3.14r2
r2=
= 100
r= 10 since r> 0
Find the rate of change when r= 10. A′(10)= 2π(10)
= 62.8
Rate of change of A when area is 314 km2 is 62.8 km2/km.
1
314 3.14
---2
16
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r km Area A km2
remember
Average rates of change are calculated using the original function, while
differentiation of this function is needed in order to calculate instantaneous rates of change at specific points.
Rates of change
1 If f(x) = x2+ 5x + 15 find:
a the average rate of change between x = 3 and x = 5
b a new function that describes the rate of change
c the (instantaneous) rate of change when x = 5.
2 A balloon is inflated so that its volume, V cm3, at any time, t seconds later is: V = − t3+ 24t2, t ∈ [0, 10]
a What is the volume of the balloon when:
i t = 0? ii t = 10?
b Hence, find the average rate of change between t = 0 and t = 10.
c Find the rate of change of volume when
i t = 0 ii t = 5 iii t = 10.
3
The average rate of change between x = 1 and x = 3 for the function y = x2 + 3x + 5 is:
4
The instantaneous rate of change of the function f(x) = x3− 3x2+ 4x, when x = −2 is:
5
If the rate of change of a function is described by = 2x2− 7x, then the function could be:
6 In a baseball game the ball is hit so that its height above the ground, h metres, is
h(t) = 1 + 18t − 3t2 t seconds after being struck.
a Find the rate of change, h′(t).
b Calculate the rate of change of height after:
i 2 seconds
ii 3 seconds
iii 4 seconds.
c What happens when t = 3 seconds?
d Find the rate of change of height when the ball first reaches a height of 16 metres.
A 1 B 9 C 5 D 3 E 7
A 2 B −2 C 28 D 3 E 12
A y = 6x3− 14x B y = x3− 7x C y = x3− x2+ 5
D y = x3− x2+ 2 E 2x2− 7x + 5
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Gradient between two points on a graph
8 5
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Gradient between two points on a graph
m
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m
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m
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dy dx ---2
3
--- 2
3 --- 7
2
---7 2
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7 The position, x metres, of a lift (above ground level) at any time, t seconds, is given by: x(t) = −2t2+ 40t
a Find the rate of change of displacement (velocity) at any time, t.
b Find the rate of change when:
i t = 5 ii t = 9 iii t = 11.
c What happened between t = 9 and t = 11?
d When and where is the rate of change zero?
8 The number of seats, N, occupied in a soccer stadium t hours after the gates are opened is given by:
N = 500t2+ 3500t, t ∈ [0, 5]
a Find N when:
i t = 1 and ii t = 3.
b What is the average rate of change between t = 1 and t = 3?
c Find the instantaneous rate when:
i t = 0 ii t = 1 iii t = 3 iv t = 4.
d Why is the rate increasing in the first 4 hours?
9 The weight, W kg, of a foal at any time, t weeks, after birth is given by: W = 80 + 12t − t2 where 0 < t < 20
a What is the weight of the foal at birth?
b Find an expression for the rate of change of weight at any time, t.
c Find the rate of change after:
i 5 weeks ii 10 weeks iii 15 weeks.
d Is the rate of change of the foal’s weight increasing or decreasing?
e When does the foal weigh 200 kg?
SkillS
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13.4
---10 The weekly profit, P (hundreds of dollars), of a factory is given by , where n is the number of employees.
a Find .
b Hence, find the rate of change of profit, in dollars per employee, if the number of employees is:
i 4
ii 16
iii 25.
c Find n when the rate of change is zero.
11 Gas is escaping from a cylinder so that its volume, V cm3, t seconds after the leak starts, is described by V= 2000 − 20t− t2.
a Find the rate of change after:
i 10 seconds
ii 50 seconds
iii 100 seconds.
b Is the rate of change ever positive? Why?
12 Assume an oil spill from an oil tanker is circular and remains that way.
a Write down a relationship between the area of the spill, A m2, and the radius,
r metres.
b Find the rate of change of A with respect to the radius, r.
c Find the rate of change of A when the radius is:
i 10 m
ii 50 m
iii 100 m.
d Is the area increasing more rapidly as the radius increases? Why?
13 A spherical balloon is being inflated.
a Express the volume of the balloon, V m3, as a function of the radius, r metres.
b Find the rate of change of V with respect to r.
c Find the rate of change when the radius is:
i 0.1 m
ii 0.2 m
iii 0.3 m.
P = 4.5n–n
dP
dn
---1 100
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14 A rectangular fish tank has a square base with its height being equal to half its base length.
a Express the length and width of the base in terms of its height, h.
b Hence, express the volume, V m3, in terms of the height, h, only.
c Find the rate of change of V when:
i h = 1 m
ii h = 2 m
iii h = 3 m.
15 For the triangular package shown find:
a x in terms of h
b the volume, V, as a function of h only
c the rate of change of V when
i h = 0.5 m
ii h = 1 m.
16 A new estate is to be established on the side of a hill.
Regulations will not allow houses to be built on slopes where the gradient is greater than 0.45. If the equation of the cross-section of the hill is
y = −0.000 02x3+ 0.006x2 find:
a the gradient of the slope
b the gradient of the slope when x equals:
i 160
ii 100
iii 40
iv 20
c the values of x where the gradient is 0.45
d the range of heights for which houses cannot be built on the hill.
17 A bushfire burns out A hectares of land, t hours after it started according to the rule A = 90t2− 3t3
a At what rate, in hectares per hour, is the fire spreading at any time, t?
b What is the rate when t equals:
i 0 ii 4 iii 8 iv 10 v 12 vi 16 vii 20?
c Briefly explain how the rate of burning changes during the first 20 hours.
d Why isn’t there a negative rate of change in the first 20 hours?
e What happens after 20 hours?
f After how long is the rate of change equal to 756 hectares per hour? 30°
h x
6
200 y
x 80
---Solving maximum and minimum problems
There are many practical situations where it is necessary to determine the maximum or minimum value of a function. For quadratic functions, differentiation makes this a rela-tively simple task. The derivative of a function gives the gradient of a tangent to the curve. When the derivative equals zero, the tangent is horizontal. The point, or points, on the curve where this occurs are called stationary points. These points can be local minimum or maximum turning points or points of inflection. A function f(x) has stationary points when f′(x) = 0.
When solving maximum or minimum problems (to obtain the value(s) of x) it should be verified that it is in fact a maximum or minimum by checking the sign of the deriv-ative to left and right of the turning point.
In the case of cubic and higher order polynomials, the local maximum (or minimum) may or may not be the highest (or lowest) value of the function in a given domain.
An example where the local maximum, found by solving f′(x) = 0, is not the largest value of f(x) in the
domain [a, b] is shown. Here, B is the point where f(x) is greatest in this domain, and is called the absolute maximum for the interval.
Case 1. The function is known
f(x) y
x
0 a b
Absolute maximum in the interval [a, b] Local
maximum
A baseball fielder throws the ball so that the equation of its path is: y = 1.5 + x − 0.02x2
where x (metres) is the horizontal distance travelled by the ball and y (metres) is the vertical height reached.
a Find the value of x for which the maximum height is reached (verify that it is a maximum).
b Find the maximum height reached.
THINK WRITE
a Write the equation of the path. a y = 1.5 + x − 0.02x2 Find the derivative . = 1 − 0.04x
Solve the equation = 0 to find the value of x for which height is a maximum.
For stationary points: = 0 1 − 0.04x = 0
−0.04x = −1 x = 25
Determine the nature of this stationary point at x = 25 by evaluating to the left and right, say, at x = 24 and at x = 26.
When x = 24
= 1 − 0.04(24)
= 0.04. When x = 26
= 1 − 0.04(26)
= −0.04.
1
2 dy
dx
--- dy
dx
---3 dy
dx
--- dy
dx
---4
dy dx
--- d---dyx Zero gradient Negative gradient Positive
gradient
dy dx
---17
Case 2. The rule for the function is not given
If the rule is not given directly then the following steps should be followed:
1. Draw a diagram if necessary and write an equation linking the given information. 2. Identify the quantity to be maximised or minimised.
3. Express this quantity as a function of one variable only (often this will be x). 4. Differentiate, set the derivative equal to zero, and solve.
5. Determine, in the case of more than one value, which one represents the maximum or minimum value.
6. For some functions, a maximum or minimum may occur at the extreme points of the domain so check these also.
7. Answer the question that is being asked.
8. Sketch a graph of the function if it helps to answer the question, noting any restric-tions on the domain.
THINK WRITE
Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 25, the stationary point is a local maximum.
The stationary point is a local maximum.
b Substitute x = 25 into y = 1.5 + x − 0.02x2 to find the corresponding y-value (maximum height).
b When x = 25,
y = 1.5 + 25 − 0.02(25)2
= 14
So the maximum height reached is 14 m.
5
A farmer wishes to fence off a rectangular paddock on a straight stretch of river so that only 3 sides of fencing are required. Find the largest possible area of the paddock if 240 metres of fencing is available.
Continued over page
THINK WRITE
Draw a diagram to represent the situation, using labels to represent the variables for length and width and write an equation involving the given information.
Let w = width l = length P = perimeter
P = l + 2w = 240 [1]
Write a rule for the area, A, of the paddock in terms of length, l, and width, w.
A = l × w [2]
1
Fence Fence
Fence w
l
w River
2
18
THINK WRITE
Express the length, l, of the rectangle in terms of the width, w, using equation [1].
l + 2w = 240
l = 240 − 2w [3]
Express the quantity, A, as a function of one variable, w, by substituting [3] into [2].
Substituting [3] into [2]: A(w) = (240 − 2w)w
= 240w − 2w2
Solve A′(w) = 0. A′(w) = 240 − 4w
For stationary points: A′(w) = 0 240 − 4w = 0
240 = 4w w = 60
Test to see if the stationary point at w = 60 will produce a maximum or minimum value for the area by evaluating A′(w) to the left and right, say, at w = 59 and at w = 61.
When w = 59 A′(59) = 240 − 4(59)
= 4. When w = 61 A′(61) = 240 − 4(61)
= −4.
Since the gradient changes from positive to negative as we move from left to right in the vicinity of w = 60, the stationary point is a local maximum.
Find the maximum area of the paddock by substituting w = 60 into the function for area.
The stationary point is a local maximum. The area of the paddock is a maximum when w = 60.
A(60) = (240 − 2 × 60) × 60
= 7200 m2
3
4
5
6
7 Zero gradient
Negative gradient Positive
gradient
8
remember
Defining a function and setting its derivative equal to zero to form an equation helps to tell us when a local maximum or minimum occurs. The solution(s) must then be substituted into the original function to find the actual maximum or minimum value(s).
Solving maximum and
minimum problems
1 A golfer hits the ball so that the equation of its path is:
y = 1.2 + x − 0.025x2 where x (metres) is the horizontal distance travelled by the ball and y (metres) is the vertical height reached.
a Find the value of x for which the maximum height is reached (and verify that it is a maximum).
b Find the maximum height reached.
2 If the volume of water, V litres, in a family’s hot water tank t minutes after the shower is turned on is given by the rule V = 200 − 1.2t2+ 0.08t3, where 0 ≤ t ≤ 15:
a find the time when the volume is minimum (that is, the length of time the shower is on)
b verify that it is a minimum by checking the sign of the derivative
c find the minimum volume
d find the value of t when the tank is full again.
3 A ball is thrown into the air so that its height, h metres, above the ground at time t seconds after being thrown is given by the function:
h(t) = 1 + 15t − 5t2
a Find the greatest height reached by the ball and the value of t for which it occurs.
b Verify that it is a maximum.
4 A gardener wishes to fence off a rectangular vegetable patch against her back fence so th
