Download Access to Engineering - Mathematics 2 ADEDEX428 Assignment 3

Access to Engineering - Mathematics 2
ADEDEX428
Assignment 3 Solutions
1. (a) Here we have to use the product rule twice.
g(x) = 4x3 e−x .
First let us differentiate
d
d −x 4x3 e−x + 4x3
e
dx
dx
= 12x2 e−x + 4x3 −e−x
= 12x2 e−x − 4x3 e−x .
g ′ (x) =
Next we differentiate h(x) = cos(−2x) ln(−5x) (where x < 0).
d
d
(cos(−2x)) ln(−5x) + cos(−2x) (ln(−5x))
dx
dx
1
= 2 sin(−2x) ln(−5x) + cos(−2x) .
x
h′ (x) =
Putting this together we obtain (for x < 0)
cos(−2x)
f ′ (x) = g ′ (x) + h′ (x) = e−x 12x2 − 4x3 + 2 sin(−2x) ln(−5x) +
.
x
(b) Again we have to use the product rule twice.
1
g(x) = x− 2 cos(2x).
First let us differentiate
1 d
d −1 x 2 cos(2x) + x− 2
(cos(2x))
dx
dx
1
1 3
= − x− 2 cos(2x) + x− 2 (−2 sin(2x))
2
1
1 −3
= − x 2 cos(2x) − 2x− 2 sin(2x).
2
g ′ (x) =
Next we differentiate h(x) = −x3 sin(−4x).
d
d
−x3 sin(−4x) + −x3
(sin(−4x))
dx
dx
= −3x2 sin(−4x) − x3 (−4 cos(−4x))
= −3x2 sin(−4x) + 4x3 cos(−4x).
h′ (x) =
Putting this together we obtain
f ′ (x) = g ′(x) + h′ (x)
1
1 3
= − x− 2 cos(2x) − 2x− 2 sin(2x) − 3x2 sin(−4x) + 4x3 cos(−4x).
2
1
(c) Here we use the quotient rule. If cos(3x) 6= 0, we have
d
d
(sin(2x)) cos(3x) − sin(2x) (cos(3x))
dx
f ′ (x) = dx
cos2 (3x)
2 cos(2x) cos(3x) − sin(2x)(−3 sin(3x))
=
cos2 (3x)
2 cos(2x) cos(3x) + 3 sin(2x) sin(3x)
.
=
cos2 (3x)
(d) Again we use the quotient rule. If x > 13 , then
d
d 3
(x ) ln(3x) − x3 (ln(3x))
dx
f ′ (x) = dx
ln(3x)2
3x2 ln(3x) − x3 x1
=
ln(3x)2
3x2 ln(3x) − x2
=
ln(3x)2
x2 (3 ln(3x) − 1)
.
=
ln(3x)2
(e) Here we have a product in the numerator and the denomiator, so we have
to use the product rule twice before we use the quotient rule. First let us
differentiate g(x) = x2 cos(3x).
d
d
g ′(x) =
x2 cos(3x) + x2 (cos(3x))
dx
dx
2
= 2x cos(3x) + x (−3 sin(3x))
= 2x cos(3x) − 3x2 sin(3x).
Next we differentiate h(x) = e2x sin(x).
d 2x d
h′ (x) =
e sin(x) + e2x (sin(x))
dx
dx
= 2e2x sin(x) + e2x cos(x)
= e2x (2 sin(x) + cos(x)).
We can now use the quotient rule. If sin(x) 6= 0, then
g ′ (x)h(x) − g(x)h′ (x)
h(x)2
(2x cos(3x) − 3x2 sin(3x)) e2x sin(x) − x2 cos(3x)e2x (2 sin(x) + cos(x))
.
=
(e2x sin(x))2
f ′ (x) =
(f) Here we just have a product in the numerator, so let us deal with that first.
If g(x) = ex (x3 − 2x2 + 5) then
d x
d
g ′(x) =
x3 − 2x2 + 5)
(e ) x3 − 2x2 + 5 + ex
dx
dx = ex x3 − 2x2 + 5 + ex 3x2 − 4x
= ex x3 + x2 − 4x + 5 .
2
Next, if h(x) = cos(−4x), then h′ (x) = 4 sin(−4x). Hence, if cos(−4x) 6= 0,
we can use the quotient rule to obtain
g ′ (x)h(x) − g(x)h′ (x)
h(x)2
ex (x3 + x2 − 4x + 5) cos(−4x) − ex (x3 − 2x2 + 5)4 sin(−4x)
=
cos2 (−4x)
x
3
2
e (x + x − 4x + 5) cos(−4x) − 4ex (x3 − 2x2 + 5) sin(−4x)
=
.
cos2 (−4x)
f ′ (x) =
(g) Here we will use the chain rule with u = x3 − 2x + 5 and f (x) = cos(u). Then
dy
dy du
=
·
= − sin(u) 3x2 − 2 = − sin x3 − 2x + 5 3x2 − 2 .
dx
du dx
(h) Here we will have to use the chain rule twice. Let us first differentiate
u(x) = sin(sin(x)). If we let v(x) = sin(x), then u(x) = sin(v). Hence
du
du dv
=
·
= cos(v) cos(x) = cos(sin(x)) cos(x).
dx
dv dx
We are now in a position to differentiate y = sin(u).
dy du
dy
=
·
= cos(u) cos(sin(x)) cos(x) = cos(sin(sin(x))) cos(sin(x)) cos(x).
dx
du dx
(i) Here we will use the chain rule with u(x) = xex and y(u) = eu . However
before we can do this, we will have to differentiate u(x) using the product
rule.
du
d
d
=
(x) ex + x (ex ) = 1ex + xex = ex (1 + x) .
dx
dx
dx
We can now use the chain rule to differentiate y(x).
dy du
dy
=
·
dx
du dx
= eu ex (1 + x)
x
= exe ex (1 + x)
= exe
x +x
(1 + x)
x(ex +1)
=e
(1 + x) .
2. 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 − 18x + 24. Thus we have to solve the equation
3x2 − 18x + 24 = 0 ⇔ x2 − 6x + 8 = 0
⇔ (x − 4)(x − 2) = 0
⇔ x = 2 or x = 4.
Thus the critical points are x = 2 and x = 4.
3
(b) In this case f ′ (x) = −6x2 + 6x + 36. Thus we have to solve the equation
−6x2 + 6x + 36 = 0 ⇔ x2 − x − 6 = 0
⇔ (x + 2)(x − 3) = 0
⇔ x = −2 or x = 3.
Thus the critical points are x = −2 and x = 3.
(c) In this case f ′ (x) = 2 cos(2x). Thus we have to solve the equation
2 cos(2x) = 0 ⇔ cos(2x) = 0
π
⇔ 2x = + kπ where k ∈ Z
2
π kπ
where k ∈ Z
⇔x= +
4
2
That is, the critical points are x =
π kπ
+
4
2
where k ∈ Z.
3. 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. Note that the critical
points of
f: R→ R
x 7→ x3 − 9x2 + 24x − 15
have already been found in Question 2(a): they are x = 2 and x = 4.
(a) In this case we evaluate f (x) at x = 0, 2, 4, 6. f (0) = −15, f (2) = 5, f (4) = 1
and f (6) = 21. Hence the global maximum of f is 21 attained at x = 6 and
the global minimum of f is −15 attained at x = 0.
(b) In this case we evaluate f (x) at x = 1, 2, 4, 6. f (1) = 1, f (2) = 5, f (4) = 1
and f (6) = 21. Hence the global maximum of f is 21 attained at x = 6 and
the global minimum of f is 1 attained at x = 1 and x = 4.
(c) In this case we evaluate f (x) at x = 0, 2, 4, 5. f (0) = −15, f (2) = 5, f (4) = 1
and f (5) = 5. Hence the global maximum of f is 5 attained at x = 2 and
x = 5 and the global minimum of f is −15 attained at x = 0.
(d) In this case we evaluate f (x) at x = 2, 4. f (2) = 5 and f (4) = 1. Hence the
global maximum of f is 5 attained at x = 2 and the global minimum of f is
1 attained at x = 4.
(e) In this case we evaluate f (x) at x = 2, 3. f (2) = 5 and f (3) = 3. Hence the
global maximum of f is 5 attained at x = 2 and the global minimum of f is
3 attained at x = 3.
(f) In this case we evaluate f (x) at x = 3, 4. f (3) = 3 and f (4) = 1. Hence the
global maximum of f is 3 attained at x = 3 and the global minimum of f is
1 attained at x = 4.
4. Note that we have found all the derivatives and critical points of these functions
in Question 2.
4
(a) Since f ′ (x) = 3x2 − 18x + 24, f ′′ (x) = 6x − 18. We now evaluate f ′′ (x) at
each of the critical points.
f ′′ (2) = 6(2)−18 = −6 < 0, so the critical point at x = 2 is a local maximum.
f ′′ (4) = 6(4) − 18 = 6 > 0, so the critical point at x = 4 is a local minimum.
(b) Since f ′ (x) = −6x2 + 6x + 36, f ′′ (x) = −12x + 6. Note it is absolutely
essential to use the correct f ′ (x) here and NOT x2 − x − 6, which we only
used in the process of solving f ′ (x) = 0.
We now evaluate f ′′ (x) at each of the critical points.
f ′′ (−2) = −12(−2) + 6 = 30 > 0, so the critical point at x = −2 is a local
minimum.
f ′′ (3) = −12(3) + 6 = −30 < 0, so the critical point at x = 3 is a local
maximum.
(c) To solve this problem we note that it appears (from looking at the graph
π
of f (x) = sin(2x)) that the maxima occur at x = + kπ and the minima
4
3π
+ kπ, where k ∈ Z. We will now prove this using the second
occur at x =
4
derivative test.
Since f ′ (x) = 2 cos(2x), f ′′ (x) = −4 sin(2x).
We now evaluate f ′′ (x) at each of the critical points.
f
′′
π
π
+ kπ = −4 sin 2
+ kπ
4
π 4
+ 2kπ
= −4 sin
2
= −4(1)
= −4
< 0,
so the critical points at x =
f
′′
π
+ kπ are local maxima.
4
3π
+ kπ
4
so the critical points at x =
3π
= −4 sin 2
+ kπ
4
3π
+ 2kπ
= −4 sin
2
= −4(−1)
=4
> 0,
3π
+ kπ are local minima.
4
5. In this question we will use the iteration formula xn+1 = xn −
(a) In this case f ′ (x) = 3x2 − 18x + 12, so that xn+1 = xn −
5
f (xn )
.
f ′ (xn )
x3 − 9x2 + 12x + 10
.
3x2 − 18x + 12
Since x0 = 2,
x30 − 9x20 + 12x0 + 10
23 − 9(22 ) + 12(2) + 10
5
x1 = x0 −
=2−
= .
2
2
3x0 − 18x0 + 12
3(2 ) − 18(2) + 12
2
Then
5
x3 − 9x2 + 12x1 + 10
= −
x2 = x1 − 1 2 1
3x1 − 18x1 + 12
2
5 3
2
2
− 9 25 + 12 52 + 10
140
.
=
2
57
3 52 − 18 52 + 12
(b) In this case f ′ (x) = −2 sin(2x) − 1, so that xn+1 = xn −
cos(2xn ) − xn
.
−2 sin(2xn ) − 1
Since x0 = 1,
x1 = x0 −
cos(2) − 1
cos(2x0 ) − x0
=1−
≃ 0.49756992.
−2 sin(2x0 ) − 1
−2 sin(2) − 1
Note that we have to use radians in this question - anytime calculus is involved, using degrees will lead to a mistake. Also note that we should keep
x1 to full calcultor accuracy. Then
cos(2x1 ) − x1
−2 sin(2x1 ) − 1
cos(0.99513984) − 0.49756992
≃ 0.49756992 −
−2 sin(0.99513984) − 1
≃ 0.515053653.
x2 = x1 −
6