Download Chapter 3 Solutions

Access to Science, Engineering and Agriculture:
Mathematics 2
MATH00040
Chapter 3 Solutions
1. (a) Since f (x) = 5 is a constant, f ′ (x) = 0.
(b) Since f (x) = −π cos(e) is a constant, f ′ (x) = 0.
(c) Here f is of the form f (x) = xn , with n = 2.
Thus f ′ (x) = 2x2−1 = 2x1 = 2x.
9
(d) Here f is of the form f (x) = xn , with n = .
2
9
9
9
7
Thus f ′ (x) = x 2 −1 = x 2 .
2
2
(e) Here f is of the form f (x) = xn , with n = −5.
Thus f ′ (x) = −5x−5−1 = −5x−6 .
(f) Here f is of the form f (x) = xn , with n = cos(2).
Thus f ′ (x) = cos(2)xcos(2)−1 .
(g) Here f is of the form f (x) = eax , with a = 4.
Thus f ′ (x) = 4e4x .
3
(h) Here f is of the form f (x) = eax , with a = .
2
3 3x
′
Thus f (x) = e 2 .
2
(i) Here f is of the form f (x) = eax , with a = −6.
Thus f ′ (x) = −6e−6x .
(j) Here f is of the form f (x) = eax , with a = π.
Thus f ′ (x) = πeπx .
(k) Here f is of the form f (x) = ln(ax), with a = 4.
1
Thus f ′ (x) = .
x
(l) Here f is of the form f (x) = ln(ax), with a = −π.
1
Thus f ′ (x) = .
x
1
(m) Here f is of the form f (x) = ln(ax), with a = .
2
1
Thus f ′ (x) = .
x
(n) Here f is of the form f (x) = sin(ax), with a = 2.
Thus f ′ (x) = 2 cos(2x).
(o) Here f is of the form f (x) = sin(ax), with a = −2.
Thus f ′ (x) = −2 cos(−2x).
1
(p) Here f is of the form f (x) = sin(ax), with a = e.
Thus f ′ (x) = e cos(ex).
(q) Here f is of the form f (x) = cos(ax), with a = 3.
Thus f ′ (x) = −3 sin(3x).
(r) Here f is of the form f (x) = cos(ax), with a = −3.
Thus f ′ (x) = −(−3) sin(−3x) = 3 sin(−3x).
(s) Here f is of the form f (x) = cos(ax), with a = −π.
Thus f ′ (x) = −(−π) sin(−πx) = π sin(−πx).
2. (a) Using the sum and multiple rules,
d
d
d
d
d
f ′ (x) =
−2x2 +
3x3 +
−4x4
(1) +
(3x) +
dx
dx
dx
dx
dx
d
d
d
d
d
x2 + 3
x3 − 4
x4
(1) + 3 (x) − 2
=
dx
dx
dx
dx
dx
2
3
= 0 + 3(1) − 2(2x) + 3(3x ) − 4(4x )
= 3 − 4x + 9x2 − 16x3 .
Note that in your assignment or exam solutions you don’t need to give as
much detail as this. I am just setting out everything carefully until you get
used to the ideas involved.
(b) Using the sum and multiple rules,
d
d
f ′ (x) =
−x−1 +
(2 sin 4x)
dx
dx
d
d
x−1 + 2 (sin 4x)
=−
dx
dx
= − −x−2 + 2(4 cos(4x))
= x−2 + 8 cos(4x).
(c) Using the sum and multiple rules,
d − 1 x
1
d
′
2
f (x) =
3e
−2 cos
x
+
dx
dx
2
d
d − 1 x
1
2
−2
=3
e
cos
x
dx
dx
2
1
1
1 −1x
x
− 2 − sin
=3 − e 2
2
2
2
1
3 −1x
= − e 2 + sin
x .
2
2
(d) Using the sum and multiple rules,
d
d
d − 3 x
′
−e 2
f (x) =
(2 ln(−x)) +
(4 cos(−3x)) +
dx
dx
dx
d
d
d − 3 x
= 2 (ln(−x)) + 4 (cos(−3x)) −
e 2
dx
dx
dx
3 −3x
1
+ 4 (3 sin(−3x)) − − e 2
=2
x
2
2
3 3
= + 12 sin(−3x) + e− 2 x .
x
2
2
(e) Using the sum and multiple rules,
d
d
d cos(1)x −2x2 +
e
(3 ln(3x)) +
dx
dx
dx
d
d cos(1)x d
x2 + 3 (ln(3x)) +
e
= −2
dx
dx
dx
1
= −2 (2x) + 3
+ cos(1)ecos(1)x
x
3
= −4x + + cos(1)ecos(1)x .
x
f ′ (x) =
(f) Using the sum and multiple rules,
d
d
d
d
(2 sin(3x)) +
(−3 sin(2x)) +
(2 cos(3x)) +
(−3 cos(2x))
dx
dx
dx
dx
d
d
d
d
= 2 (sin(3x)) − 3 (sin(2x)) + 2 (cos(3x)) − 3 (cos(2x))
dx
dx
dx
dx
= 2 (3 cos(3x)) − 3 (2 cos(2x)) + 2 (−3 sin(3x)) − 3 (−2 sin(2x))
= 6 cos(3x) − 6 cos(2x) − 6 sin(3x) + 6 sin(2x)
= 6(cos(3x) − cos(2x) − sin(3x) + sin(2x)).
f ′ (x) =
(g) Using the sum rule,
f ′ (x) =
d 2
d 2x e −4 +
e
= 0 + 2e2x = 2e2x .
dx
dx
Note that here we didn’t need the multiple rule and also we were able to deal
with the two terms e2 and −4 all at once since e2 − 4 is just a constant.
(h) Using the sum and multiple rules,
d
d
d
−3x−3 +
4x4 +
5x−5 +
dx
dx
dx
d
d
d
x−3 + 4
x4 + 5
x−5 +
= −3
dx
dx dx −4
3
= −3 −3x
+ 4 4x + 5 −5x−6 + 0
= 9x−4 + 16x3 + −25x−6 .
f ′ (x) =
d
(3)
dx
d
(3)
dx
Note that 3x0 is just the number 3 (unless x = 0 when x0 is not defined), so
it differentiates to zero. We could also obtain the derivative as 3 (0x−1 ) = 0
but it would be a bit more work.
For the remainder of the questions we will still need the sum and multiple rules
but I will not mention them explicitly.
3. (a) Here we have to use the product rule twice.
g(x) = 3x2 sin(2x).
First let us differentiate
d
d
3x2 sin(2x) + 3x2 (sin(2x))
dx
dx
2
= 6x sin(2x) + 3x (2 cos(2x))
= 6x sin(2x) + 6x2 cos(2x).
g ′ (x) =
3
Next we differentiate h(x) = −e2x cos(x).
d
d
−e2x cos(x) − e2x (cos(x))
dx
dx
2x
2x
= −2e cos(x) − e (− sin(x))
= −2e2x cos(x) + e2x sin(x).
h′ (x) =
Putting this together we obtain
f ′ (x) = g ′(x) + h′ (x) = 6x sin(2x) + 6x2 cos(2x) − 2e2x cos(x) + e2x sin(x).
(b) Again we have to use the product rule twice.
g(x) = (2x2 − 3x−3 + 5x4 ) sin(−5x).
First let us differentiate
d
d
2x2 − 3x−3 + 5x4 sin(−5x) + (2x2 − 3x−3 + 5x4 ) (sin(−5x))
dx
dx
−4
3
2
−3
4
= 4x + 9x + 20x sin(−5x) + (2x − 3x + 5x ) (−5 cos(−5x))
= 4x + 9x−4 + 20x3 sin(−5x) − 5(2x2 − 3x−3 + 5x4 ) cos(−5x).
g ′ (x) =
Next we differentiate h(x) = e−3x (sin(2x) − cos(x)).
d
d −3x e
(sin(2x) − cos(x)) + e−3x (sin(2x) − cos(x))
dx
dx
−3x
−3x
= −3e (sin(2x) − cos(x)) + e (2 cos(2x) + sin(x)).
h′ (x) =
Putting this together we obtain
f ′ (x) =g ′ (x) + h′ (x)
=(4x + 9x−4 + 20x3 ) sin(−5x) − 5(2x2 − 3x−3 + 5x4 ) cos(−5x)
− 3e−3x (sin(2x) − cos(x)) + e−3x (2 cos(2x) + sin(x)).
(c) Here there is a quick method and a slow method.
The quick method is to spot
f (x) = sin(x) sin(x) + cos(x) cos(x) = sin2 (x) + cos2 (x) = 1,
so that f ′ (x) = 0.
For the slow method, we use the product rule twice:
d
d
(sin(x)) sin(x) + sin(x) (sin(x))
dx
dx
d
d
+
(cos(x)) cos(x) + cos(x) (cos(x))
dx
dx
= cos(x) sin(x) + sin(x) cos(x) − sin(x) cos(x) + cos(x)(− sin(x))
=0, as expected.
f ′ (x)) =
Note that when we study the chain rule we will see another way of differentiating functions such as sin2 (x) and cos2 (x).
4
(d) Here there are also two methods.
For the first method we will use the half angle trig formulae to obtain
f (x) = sin2 (x) − cos2 (x) =
1 − cos(2x) 1 + cos(2x)
−
= − cos(2x),
2
2
so that f ′ (x) = 2 sin(2x).
For the second method we use the product rule twice:
d
d
(sin(x)) sin(x) + sin(x) (sin(x))
dx dx
d
d
−
(cos(x)) cos(x) + cos(x) (cos(x))
dx
dx
= cos(x) sin(x) + sin(x) cos(x) − [− sin(x) cos(x) + cos(x)(− sin(x))]
=4 sin(x) cos(x).
f ′ (x)) =
This ‘looks’ different from our first answer but it is in fact the same, since
using the double angle trig formula, we obtain
2 sin(2x) = 2(2 sin(x) cos(x)) = 4 sin(x) cos(x).
This illustrates an important general point. If you use two different methods
to solve a problem, you might end up with answers that look different but
are in fact the same.
(e) Here we just have to use one application of the product rule to differentiate
g(x) = e2x ln(2x).
g ′ (x)) =
d 2x d
e ln(2x) + e2x (ln(x)))
dx
dx
1
= 2e2x ln(2x) + e2x
x
2x
e
.
= 2e2x ln(2x) +
x
Since the derivative of h(x) = −x2 + x3 + 1 is h′ (x) = −2x + 3x2 , we have
f ′ (x) = g ′(x) + h′ (x) = 2e2x ln(2x) +
e2x
− 2x + 3x2 .
x
(f) Here we don’t need to use the product rule at all. Since e0 = 1 and ln(1) = 0,
we have f (x) = cos(2x), so that f ′ (x) = −2 sin(2x).
It is very important to keep your eye out for situations like this, where an
initial simplification makes the differentiation a lot easier.
(g) Here we have a product of three terms, so things are a little more complicated.
What we will do is to use the product rule to differentiate a product two of
the terms and then use the product rule again.
First we will differentiate g(x) = x2 e3x .
g ′ (x) =
d 3x d
x2 e3x + x2
e
= 2xe3x + x2 3e3x = 2xe3x + 3x2 e3x .
dx
dx
5
We can now differentiate f by writing it as f (x) = (x2 e3x )(cos(4x)).
d
d
x2 e3x cos(4x) + x2 e3x (cos(4x))
dx
dx
3x
2 3x
= 2xe + 3x e cos(4x) + x2 e3x (−4 sin(4x))
= (2xe3x + 3x2 e3x ) cos(4x) − 4x2 e3x sin(4x).
f ′ (x) =
4. (a) Here we use the quotient rule. If sin(x) 6= 0, we have
d
d
(x) sin(x) − x (sin(x))
sin(x) − x cos(x)
dx
=
.
f ′ (x) = dx
2
sin (x)
sin2 (x)
(b) By the quotient rule, we have
d
d
(2e3x ) (x2 + x + 1) − 2e3x (x2 + x + 1)
dx
f ′ (x) = dx
(x2 + x + 1)2
6e3x (x2 + x + 1) − 2e3x (2x + 1)
=
.
(x2 + x + 1)2
(c) In this case we will first have to use the product rule to differentiate
g(x) = x cos(2x) and then use the quotient rule. Now
d
d
(x) cos(2x) + x (cos(2x))
dx
dx
= 1 cos(2x) + x (−2 sin(2x))
= cos(2x) − 2x sin(2x).
g ′ (x) =
We can now use the quotient rule to obtain (when x > 12 )
d
d
(x cos(2x)) ln(2x) − x cos(2x) (ln(2x))
dx
f ′ (x) = dx
(ln(2x))2
(cos(2x) − 2x sin(2x)) ln(2x) − x cos(2x) x1
=
(ln(2x))2
(cos(2x) − 2x sin(2x)) ln(2x) − cos(2x)
.
=
(ln(2x))2
(d) Here we will have to use the product rule for the numerator and the denomonator before we use the quotient rule.
First we will differentiate g(x) = (x3 − 2x) cos(−2x) using the product rule.
d
d
x3 − 2x cos(−2x) + (x3 − 2x) (cos(−2x))
dx
dx
= 3x2 − 2 cos(−2x) + (x3 − 2x) (2 sin(−2x))
= 3x2 − 2 cos(−2x) + 2(x3 − 2x) sin(−2x).
g ′(x) =
6
Next we will differentiate h(x) = sin(x) cos(x) using the product rule. Note
we could also use sin(x) cos(x) = 21 sin(2x) in which case the answer below
would look different but would be the same.
d
d
(sin(x)) cos(x) + sin(x) (cos(x))
dx
dx
d
= cos(x) cos(x) + sin(x) (− sin(x))
dx
2
2
= cos (x) − sin (x).
h′ (x) =
We can now use the quotient rule to differentiate f (when sin(x) cos(x) 6= 0).
d
((x3 − 2x) cos(−2x)) sin(x) cos(x)
′
dx
f (x) =
(sin(x) cos(x))2
d
(x3 − 2x) cos(−2x) (sin(x) cos(x))
dx
−
(sin(x) cos(x))2
((3x2 − 2) cos(−2x) + 2(x3 − 2x) sin(−2x)) sin(x) cos(x)
=
sin2 (x) cos2 (x)
((x3 − 2x) cos(−2x))(cos2 (x) − sin2 (x))
−
sin2 (x) cos2 (x)
(e) By the quotient rule, we have (when sin(x) 6= 0)
d
d
(cos(x)) sin(x) − cos(x) (sin(x))
dx
f ′ (x) = dx
2
(sin(x))
− sin(x) sin(x) − cos(x) cos(x)
=
sin2 (x)
1
=− 2
(using sin2 (x) + cos2 (x) = 1)
sin (x)
= − cosec2 (x).
(f) Before we use the quotient rule here, we will have to use the product rule to
differentiate g(x) = 2 cos(3x) ln(3x). When x > 0,
d
d
(2 cos(3x)) ln(3x) + 2 cos(3x) (ln(3x))
dx
dx
1
= −6 sin(3x) ln(3x) + 2 cos(3x)
x
2 cos(3x)
= −6 sin(3x) ln(3x) +
.
x
g ′(x) =
7
Now using the quotient rule, when x > 0, we have
d
(2 cos(3x) ln(3x) + x2 ) (2e−x + x4 )
′
dx
f (x) =
(2e−x + x4 )2
d
(2e−x + x4 )
(2 cos(3x) ln(3x) + x2 )
dx
−
(2e−x + x4 )2
2 cos(3x)
+ 2x (2e−x + x4 )
−6 sin(3x) ln(3x) +
x
=
(2e−x + x4 )2
(2 cos(3x) ln(3x) + x2 )(−2e−x + 4x3 )
.
−
(2e−x + x4 )2
5. (a) Here we will use the chain rule with u = x2 and y = f (x) = eu . Then
f ′ (x) =
dy
dy du
2
=
·
= eu (2x) = 2xex .
dx
du dx
(b) Here we will use the chain rule with u = 2x2 + 3x and y = f (x) = sin(u).
Then
f ′ (x) =
dy
dy du
=
·
= cos(u) (4x + 3) = (4x + 3) cos 2x2 + 3x .
dx
du dx
(c) Here we will use the chain rule with u = sin(x) and y = f (x) = ln(u). Then
f ′ (x) =
dy
dy du
1
cos(x)
=
·
= cos(x) =
= cot(x).
dx
du dx
u
sin(x)
(d) Before we use the chain rule here, we will first use the product rule to differentiate u = xex . We have
d
d
du
=
(x) ex + x (ex ) = (1)ex + xex = ex + xex .
dx
dx
dx
We can now use the chain rule with u = xex and y = f (x) = cos(u). Then
f ′ (x) =
dy
dy du
=
·
= − sin(u) (ex + xex ) = −(ex + xex ) sin (xex ) .
dx
du dx
(e) Here we will use the chain rule with u = sin(x) and y = f (x) = sin(u). Then
f ′ (x) =
dy
dy du
=
·
= cos(u) cos(x) = cos(sin(x)) cos(x) = cos(x) cos(sin(x)).
dx
du dx
(f) Here we will use the chain rule twice. We will first use it to differentiate
u = sin(cos(x)). To do this we let v = cos(x) and u = sin(v). Then
du
du dv
=
·
= cos(v)(− sin(x)) = − sin(x) cos(v) = − sin(x) cos(cos(x)).
dx
dv dx
8
Next we will use the chain rule again, this time with u = sin(cos(x)) and
y = f (x) = cos(u).
dy
dx
dy du
·
=
du dx
= − sin(u)(− sin(x) cos(cos(x)))
= sin(sin(cos(x))) sin(x) cos(cos(x))
= sin(x) cos(cos(x)) sin(sin(cos(x))).
f ′ (x) =
(g) Again we will use the chain rule twice. We will first use it to differentiate
u = sin(x2 ). To do this we let v = x2 and u = sin(v). Then
du
du dv
=
·
= cos(v)(2x) = 2x cos(x2 ).
dx
dv dx
We will now use the chain rule again, this time with u = sin(x2 ) and
y = f (x) = eu .
f ′ (x) =
dy du
dy
2
=
·
= eu (2x cos(x2 )) = 2x cos(x2 )esin(x ) .
dx
du dx
6. In all these questions we will differentiate f (x) and solve the equation f ′ (x) = 0
to find the critical points.
(a) In this case f ′ (x) = 3x2 − 6x − 9. Thus we have to solve the equation
3x2 − 6x − 9 = 0 ⇔ x2 − 2x − 3 = 0
⇔ (x + 1)(x − 3) = 0
⇔ x = −1 or x = 3.
Thus the critical points are x = −1 and x = 3.
(b) In this case f ′ (x) = 3x2 + 6x. Thus we have to solve the equation
3x2 + 6x = 0 ⇔ x2 + 2x = 0
⇔ (x + 2)x = 0
⇔ x = −2 or x = 0.
Thus the critical points are x = −2 and x = 0.
(c) In this case f ′ (x) = −6x2 − 18x + 24. Thus we have to solve the equation
−6x2 − 18x + 24 = 0 ⇔ x2 + 3x − 4 = 0
⇔ (x + 4)(x − 1) = 0
⇔ x = −4 or x = 1.
Thus the critical points are x = −4 and x = 1.
9
(d) In this case f ′ (x) = 6x2 + 6x + 6. Thus we have to solve the equation
6x2 + 6x + 6 = 0 ⇔ x2 + x + 1 = 0.
However the equation x2 + x + 1 = 0 has no real solutions, so there are no
critical points.
(e) In this case f ′ (x) = cos(x). Thus we have to solve the equation
cos(x) = 0 ⇔ x =
π
+ kπ for each k ∈ Z.
2
π
+ kπ for each k ∈ Z.
2
(f) In this case f ′ (x) = −2 sin(2x). Thus we have to solve the equation
Thus the critical points are x =
−2 sin(2x) = 0 ⇔ sin(2x) = 0
⇔ 2x = kπ for each k ∈ Z
kπ
⇔x=
for each k ∈ Z
2
Thus the critical points are x =
kπ
for each k ∈ Z.
2
7. In these questions we will find where the global maxima and minima of each of
the following functions occur by evaluating the functions at the endpoints of the
domain and at any critical points that lie in the domain.
(a) Note that the critical points of
f: R→ R
x 7→ x3 − 3x2 − 9x + 12
have already been found in Question 6(a): they are x = −1 and x = 3.
(i) In this case we evaluate f (x) at x = −4, −1, 3, 6. f (−4) = −64,
f (−1) = 17, f (3) = −15 and f (6) = 66.
Hence the global maximum of f is 66 attained at x = 6 and the global
minimum of f is −64 attained at x = −4.
(ii) In this case we evaluate f (x) at x = −3, −1, 3, 5. f (−3) = −15,
f (−1) = 17, f (3) = −15 and f (5) = 17.
Hence the global maximum of f is 17 attained at x = −1 and x = 5,
and the global minimum of f is −15 attained at x = −3 and x = 3.
(iii) In this case we evaluate f (x) at x = −2, −1, 3, 4. f (−2) = 10,
f (−1) = 17, f (3) = −15 and f (4) = −8.
Hence the global maximum of f is 17 attained at x = −1 and the global
minimum of f is −15 attained at x = 3.
(iv) In this case we evaluate f (x) at x = 0, 3. f (0) = 12 and f (3) = −15.
Hence the global maximum of f is 12 attained at x = 0 and the global
minimum of f is −15 attained at x = 3.
10
(b) Note that the critical points of
f: R→R
x 7→ x3 + 3x2 − 5
have already been found in Question 6(b): they are x = −2 and x = 0.
(i) In this case we evaluate f (x) at x = −3, −2, 0, 1. f (−3) = −5,
f (−2) = −1, f (0) = −5 and f (1) = −1.
Hence the global maximum of f is −1 attained at x = −2 and x = 1,
and the global minimum of f is −5 attained at x = −3 and x = 0.
(ii) In this case we evaluate f (x) at x = −3, −2, 0. f (−3) = −5,
f (−2) = −1 and f (0) = −5.
Hence the global maximum of f is −1 attained at x = −2 and the
global minimum of f is −5 attained at x = −3 and x = 0.
(iii) In this case we evaluate f (x) at x = −2, 0, 1. f (−2) = −1, f (0) = −5
and f (1) = −1.
Hence the global maximum of f is −1 attained at x = −2 and x = 1,
and the global minimum of f is −5 attained at x = 0.
(iv) In this case we evaluate f (x) at x = −1, 0. f (−1) = −3 and f (0) = −5.
Hence the global maximum of f is −3 attained at x = −1 and the global
minimum of f is −5 attained at x = 0.
(c) Note that the critical points of
f: R→ R
x 7→ −2x3 − 9x2 + 24x − 1
have already been found in Question 6(c): they are x = −4 and x = 1.
(i) In this case we evaluate f (x) at x = −7, −4, 1, 3. f (−7) = 76,
f (−4) = −113, f (1) = 12 and f (3) = −64.
Hence the global maximum of f is 76 attained at x = −7 and the global
minimum of f is −113 attained at x = −4.
(ii) In this case we evaluate f (x) at x = −6, −4, 1, 4. f (−6) = −37,
f (−4) = −113, f (1) = 12 and f (4) = −177.
Hence the global maximum of f is 12 attained at x = 1 and the global
minimum of f is −177 attained at x = 4.
(iii) In this case we evaluate f (x) at x = −4, 1. f (−4) = −113 and
f (1) = 12.
Hence the global maximum of f is 12 attained at x = 1 and the global
minimum of f is −113 attained at x = −4.
(iv) In this case we evaluate f (x) at x = 2, 3. f (2) = −5 and
f (3) = −64.
Hence the global maximum of f is −5 attained at x = 2 and the global
minimum of f is −64 attained at x = 3.
11
(d) Note that we saw in Question 6(d) that the function
f: R→ R
x 7→ −2x3 − 9x2 + 24x − 1
has no critical points.
(i) In this case we evaluate f (x) at x = −1, 1. f (−1) = 0 and f (1) = 16.
Hence the global maximum of f is 16 attained at x = 1 and the global
minimum of f is 0 attained at x = −1.
(ii) In this case we evaluate f (x) at x = −1, 3. f (−1) = 0 and f (3) = 104.
Hence the global maximum of f is 104 attained at x = 3 and the global
minimum of f is 0 attained at x = −1.
(iii) In this case we evaluate f (x) at x = −4, 1. f (−4) = −99 and
f (1) = 12.
Hence the global maximum of f is 12 attained at x = 1 and the global
minimum of f is −99 attained at x = −4.
(iv) In this case we evaluate f (x) at x = −4, 3. f (−4) = −99 and
f (3) = 104.
Hence the global maximum of f is 104 attained at x = 3 and the global
minimum of f is −99 attained at x = −4.
(e) Note that the critical points of
f: R→R
x 7→ sin(x)
π
have already been found in Question 6(e): they are x =
+ kπ for each
2
k ∈ Z.
π
3π
+ 2kπ = 1 and sin
+ 2kπ = −1 for each k ∈ Z,
(i) Since sin
2
2
π
the global maximum of f is 1 attained at x = + 2kπ for each k ∈ Z,
2
3π
+ 2kπ for each
and the global minimum of f is −1 attained at x =
2
k ∈ Z.
π
π π
1
(ii) In this case we evaluate f (x) at x = − , . f −
= − √ and
4 4
4
2
π 1
=√ .
f
4
2
1
π
Hence the global maximum of f is √ attained at x = and the global
4
2
π
1
minimum of f is − √ attained at x = − .
4
2
π π 3π
= 1,
(iii) In this case we evaluate f (x) at x = 0, , , 2π. f (0) = 0, f
2 2
2
3π
f
= −1 and f (2π) = 0.
2
π
Hence the global maximum of f is 1 attained at x = and the global
2
3π
.
minimum of f is −1 attained at x =
2
12
(f) Note that the critical points of
f: R→R
x 7→ cos(2x)
kπ
for each k ∈ Z.
have already been found in Question 6(f): they are x =
2
π
(i) Since cos (2(kπ)) = 1 and cos 2
+ kπ
= cos(π + 2kπ) = −1 for
2
each k ∈ Z, the global maximum of f is 1 attained at x = kπ for each
π
k ∈ Z, and the global minimum of f is −1 attained at x = + kπ for
2
each k ∈ Z.
π
π
(ii) In this case we evaluate f (x) at x = − , 0, .
4
4
π
π
= cos −
= 0, f (0) = cos(0) = 1 and
f −
π 4
π 2
f
= cos
= 0.
4
2
Hence the global maximum of f is 1 attained at x = 0 and the global
π
π
minimum of f is 0 attained at x = − and x = .
4
4
π
3π
(iii) In this case we evaluate f (x) at x = 0, , π, , 2π. f (0) = cos(0) = 1,
2
2 π 3π
= cos(π) = −1, f (π) = cos(2π) = 1, f
= cos(3π) = −1,
f
2
2
and f (2π) = cos(4π) = 1.
Hence the global maximum of f is 1 attained at x = 0, x = π and
π
x = 2π, and the global minimum of f is −1 attained at x =
and
2
3π
x=
.
2
The global maxima occur at (0,
1), (π,1) and (2π, 1) and the global
π
3π
, −1 and
, −1 .
minima occur at
2
2
8. Note that we have found all the derivatives and critical points of these functions
in Question 6.
(a) Since f ′ (x) = 3x2 − 6x − 9, f ′′ (x) = 6x − 6.
We now evaluate f ′′ (x) at each of the critical points.
f ′′ (−1) = 6(−1) − 6 = −12 < 0, so the critical point at x = −1 is a local
maximum.
f ′′ (3) = 6(3) − 6 = 12 > 0, so the critical point at x = 3 is a local minimum.
(b) Since f ′ (x) = 3x2 + 6x, f ′′ (x) = 6x + 6.
We now evaluate f ′′ (x) at each of the critical points.
f ′′ (−2) = 6(−2) + 6 = −6 < 0, so the critical point at x = −2 is a local
maximum.
f ′′ (0) = 6(0) + 6 = 6 > 0, so the critical point at x = 0 is a local minimum.
(c) Since f ′ (x) = −6x2 − 18x + 24, f ′′ (x) = −12x − 18.
We now evaluate f ′′ (x) at each of the critical points.
f ′′ (−4) = −12(−4) − 18 = 30 > 0, so the critical point at x = −4 is a local
13
minimum.
f ′′ (1) = −12(1) − 18 = −30 < 0, so the critical point at x = 1 is a local
maximum.
(d) The function f ′ (x) = 6x2 + 6x + 6 has no critical points to classify.
(e) Since f ′ (x) = cos(x), f ′′ (x) = − sin(x).
Wenow evaluate
f ′′ (x)at each of
the critical points.
π
π
+ 2kπ = − sin
+ 2kπ = −1 < 0, so the critical points at
f ′′
2
2
π
x = + 2kπ are local maxima for each k ∈ Z.
2
3π
3π
′′
f
+ 2kπ = − sin
+ 2kπ = −(−1) = 1 > 0, so the critical
2
2
3π
+ 2kπ are local minima for each k ∈ Z.
points at x =
2
(f) Since f ′ (x) = −2 sin(2x), f ′′ (x) = −4 cos(2x).
We now evaluate f ′′ (x) at each of the critical points.
f ′′ (kπ) = −4 cos (2kπ) = −4(1) = −4 < 0, so the critical points at x = kπ
arelocal maxima
for each k ∈ Z.
′′ π
f
+ kπ = −4 cos (π + 2kπ) = −4(−1) = 4 > 0, so the critical points at
2
π
x = + kπ are local minima for each k ∈ Z.
2
9. In this question we will use the iteration formula xn+1 = xn −
(a) In this case f ′ (x) = 3x2 − 6x − 9, so that xn+1 = xn −
Since x0 = 4,
x1 = x0 −
f (xn )
.
f ′ (xn )
x3n − 3x2n − 9xn + 12
.
3x2n − 6xn − 9
x30 − 3x20 − 9x0 + 12
43 − 3(42 ) − 9(4) + 12
68
=
4
−
= .
2
2
3x0 − 6x0 − 9
3(4 ) − 6(4) − 9
15
Then
x31 − 3x21 − 9x1 + 12
3x21 − 6x1 − 9
68 2
68 3
− 3 15
− 9 68
+ 12
68
15
15
=
−
2
68
15
3 68
− 6 15
−9
15
x2 = x1 −
= 4.4268 to 4 d.p.
(b) In this case f ′ (x) = 3x2 + 6x, so that xn+1 = xn −
x3n + 3x2n − 5
.
3x2n + 6xn
Since x0 = 1,
x1 = x0 −
x30 + 3x20 − 5
13 + 3(12) − 5
10
=
1
−
= .
2
2
3x0 + 6x0
3(1 ) + 6(1)
9
Then
10
x3 + 3x2 − 5
=
−
x2 = x1 − 1 2 1
3x1 + 6x1
9
14
2
10 3
+ 3 10
−
9
9
2
10
3 10
+
6
9
9
5
= 1.1038 to 4 d.p.
(c) Here f ′ (x) = −6x2 − 18x + 24, so that xn+1 = xn −
Since x0 = 2,
x1 = x0 −
−2x3n − 9x2n + 24xn − 12
.
−6x2n − 18xn + 24
−2x30 − 9x20 + 24x0 − 12
−2(23 ) − 9(22 ) + 24(2) − 12
14
=
2
−
= .
2
2
−6x0 − 18x0 + 24
−6(2 ) − 18(2) + 24
9
Then
−2x31 − 9x21 + 24x1 − 12
−6x21 − 18x1 + 24
3
14 2
14
− 12
−
9
+
24
14 −2 14
9
=
−
2 9
9
14
14
9
−6 9 − 18 9 + 24
x2 = x1 −
= 1.3410 to 4 d.p.
(d) Here f ′ (x) = − sin(x) − 1, so that xn+1 = xn −
Since x0 = 0,
x1 = x0 −
cos(xn ) − xn
.
− sin(xn ) − 1
cos(0) − 0
cos(x0 ) − x0
=0−
= 1.
− sin(x0 ) − 1
− sin(0) − 1
Then
x2 = x1 −
cos(x1 ) − x1
cos(1) − 1
=1−
= 0.7504 to 4 d.p.
− sin(x1 ) − 1
− sin(1) − 1
15