Download MA11001 Calculus Workshop 4 Week 8, Answers

MA11001 Calculus
Workshop 4 Week 8, Answers
1. Differentiate the following functions and simplify where possible.
a.
= 2cos2x
d
dx
d
dx
sin 2x
d
dx
d
dx
d
dx
sinÝ4x 2 + 3Þ = 8 cos 4x 2 + 3
2sinÝ x Þ cosÝxÞ = 4cos 2 x ? 2 = 2cos2x (since 2sinÝ x Þ cosÝxÞ = sin2xÞ
cos2 x
c. d sin 2 x =
dx
x
d. d 4sin 2x + cos 3x
= 8cos2x ? 3sin3x
dx
b.
e.
f.
g.
sin
3
4x
= 12sin
2
4x cos 4x
3x 2 ? 4cos 1 ? 2x
= 6x + 8sin ?1 + 2x
+ cos 3x
=
h.
d
dx
i.
d
dx
d
dx
tanÝ2x ? 4Þ = 2sec 2 2x ? 4
k.
d
dx
sec
l.
d
dx
sinÝ x Þ secÝ x Þ =
j.
4sin 2 x
sin 3 Ý x Þ + sinÝ x Þ cos 2 Ý x Þ
x?4
= 1
2
d
dx
x
sec
=
x?4
4cos2 x ? 3 sin3x
x
x
d
dx
sinÝ x Þ = cosÝ x Þ since sin 2 x + cos 2 x = 1.
tan
x?4
x?4
tanÝ x Þ = sec 2 Ý x Þ .
2. Find the second, third and fourth derivatives of functions fÝ x Þ = sinÝ ^ + x Þ and gÝ x Þ = sin 2x (See Ý a Þ and
b
in Question 1). Can you identify a pattern that emerges ? What are the nth derivatives (n 5 NÞ?
sinÝ ^ + x Þ = sinÝ ^ Þ cosÝ x Þ + cosÝ ^ Þ sinÝ x Þ = ? sinÝ x Þ .
f v Ý x Þ = ? cosÝ x Þ , f vv Ý x Þ = sinÝ x Þ ,f vvv Ý x Þ = cosÝ x Þ . The fourth derivative returns us to the original function:
4
Ý x Þ = fÝ x Þ
and then higher derivatives cycle through the same sequence of 4 functions.
g v Ý x Þ = 2cos 2x , g vv Ý x Þ = ?4sin 2x , g vvv Ý x Þ = ?8cos 2x and the fourth derivative returns us to 2 4 times
the original function:
f
g
3. Let y = 4sin 2x
4
Ý x Þ = 2 4 gÝ x Þ
? 5cos 2x . Find the second derivative of y and find a constant k so that
d 2 y = ?k 2 y.
dx 2
Also show that
d 4 y = k 4 y.
dx 4
v
vv
y = 8cos2x + 10sin2x, y = ?16sin2x + 20cos2x = ?4y. Hence k = 2.
Since y vv = 4y, differentiating twice more gives
y vvvv = 4y vv = 16y.
4. Let y = tanÝ x Þ . Find the second derivative of y and show that
d 2 y = 2y 1 + y 2 .
dx 2
v
2
2
2
y = sec x and using 1 + tan x = sec x we find that
yv = 1 + y2 .
Differentiating again gives (using the chain rule on y 2 Þ
y vv = 2yy v = 2y 1 + y 2 .
5. In each case sketch the graph of the function f and use the horizontal line test to determine whether the
function f is one-to-one.
a. fÝxÞ = 3x + 2 is one-to-one.
1
8
6
4
2
-2
0
-1
1x
2
-2
-4
b. fÝxÞ =
x ? 1 : maximal domain is x 5 ß1,KÞ is one-to-one.
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
1
2
3
x
4
5
c. fÝxÞ = |x |, x 5 R is not one-to-one.
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
-2
0
-1
1x
2
d. fÝxÞ = x 2 ? 2x + 2, x 5 R is not one–to–one.
5
4
3
2
1
-1
0
1
2
x
3
e. fÝxÞ = x 3 ? 3x is not one–to–one.
15
10
5
-3
-2
-1
0
1
x
2
3
-5
-10
-15
f. fÝxÞ = x|x |, x 5 R is one–to–one.
2
4
2
-2
-1
0
x1
2
-2
-4
6. In each case determine whether the function f is one-to-one by investigating the sign of f v ÝxÞ.
a. fÝxÞ = x 2 + 8x + 7, f v Ý x Þ = 2x + 8 changes sign at x = ?4, so f cannot be one–to–one.
b. fÝxÞ = 5x 7 + 3x 3 + 17x ? 9, f v Ý x Þ = 35x 6 + 9x 2 + 17 > 0 and so f is an increasing function—it must be
one–to–one.
c. fÝxÞ = 2x + sinÝxÞ, f v Ý x Þ = 2 + cosÝ x Þ ³ 1 so f is an increasing function—it must be one–to–one.
x , f v Ý x Þ = 1 > 0 so f is an increasing function—it must be one–to–one.
x+1 2
x+1
7. Show that the function fÝxÞ = 3 ? x is its own inverse. One possible test is whether fÝ x Þ = f ?1 Ý x Þ . What is
1?x
another way of testing the assertion? Look in the lecture notes for properties of inverse functions.
Solving y = fÝ x Þ for x :
y= 3?x ì y 1?x = 3?x
1?x
x ? xy = 3 ? y ì x = 3 ? y = fÝ y Þ
1?y
?1
So, since x = fÝ y Þ , f Ý y Þ = fÝ y Þ and f is its own inverse.
Alternatively: we could use the proprty of inverse functions:
d. fÝxÞ =
f f ?1 Ý x Þ
= x and f ?1 fÝ x Þ
= x.
To show that f is its own inverse, we have to compute
f fÝ x Þ = 3 ? u ,
u = fÝ x Þ = 3 ? x
1?u
1?x
3
?
x
3?
1?x = x
f fÝ x Þ =
1? 3?x
1?x
thus verifying that f is its own inverse.
8. Find conditions on a,b,c,d which ensure that the function fÝxÞ = ax + b is its own inverse.
cx + d
f fÝ x Þ = au + b ,
u = ax + b
cu + d
cx + d
ax
+
b
a
+b
2
= cx + d
= a x + ab + bcx + bd2
cax + cb + dcx + d
c ax + b + d
cx + d
so f fÝ x Þ = x when
ab + bd = 0 é b a + d
=0
ac + cd = 0 é c a + d
=0
a + bc = bc + d é a = d 2 .
2
so either d = ?a or
2
2
b = c = 0 and a = d
9. Show that the function fÝxÞ = x 3 ? 3x 2 + 2x is not one-to-one on Ý?K,+KÞ.
f v Ý x Þ = 3x 2 ? 6x + 2 = 3 x ? 1 2 ? 1 so that f v Ý x Þ changes sign when x = 1 ± 13 . Hence f cannot be
one–to–one.
Find the largest value of k so that f is one-to-one on the interval Ý?k,kÞ.
f is one–to–one on ?k,k where k = 23 .
3
4
3
2
1
-2
0
-1
1
x
2
3
-1
-2
-3
-4
10. What is the largest value of k so that f Ý x Þ = x 3 ? 3kx 2 + 6x is one-to-one on the interval Ý?K,+KÞ?
ÝHint: complete the square in the derivative of f.Þ
f v Ý x Þ = 3x 2 ? 6kx + 6 = 3 x ? k 2 ? 3k 2 + 6 so f v Ý x Þ ³ 0 when ?3k 2 + 6 ³ 0, i.e. k ² 2 so the largest value
of k is 2.
11. Apply the same ideas to differentiate
(b) cos x ? 3
(a) sin 2 3x + 5
(c) tan
4
1?x
(d) cos 2 x ? 3
2
(e) sec ax
x + sin
(g)
d
dx
sin 2 3x + 5
b.
d
dx
cos x ? 3 = ? 1
2
4
1?x
2
1
x
sin 2 2x + 1
(i)
a.
(f)
+ cos 2 2x + 1
d
dx
tan
d.
d
dx
cos 2 x ? 3 = ?
h.
x+1
x?1
3
.
= 3sin 6x + 10
x?3
cos
sec 2 ?1 + x 2
sin
x?3
x?3
= ? 1
2
x?3
sec Ý ax Þ tan Ý ax Þ
sec ax = a
2
Ý ax Þ
1
cos x
f. d sin 1x = ?
dx
x2
x 3 ? xcos 1x +
2
g. d x + sin 1x
=
2
dx
e.
cos
x?3
sin
= 8xtan 3 ?1 + x 2
c.
(j)
1
x
5 ? 1 ? x2
(h)
= 6sin 3x + 5 cos 3x + 5
2
sin
x?3
sin2
x?3
d
dx
d
dx
d
dx
j.
d
dx
sin 2 2x + 1
cos
x+1
x?1
3
x 2 ? sin 1x cos 1x
x2
1
5 ? 1 ? x2 =
5?
2
i.
sin 1x
1?x
x
1?x
2
+ cos 2 2x + 1
= 6 sin
2
= 0 since sin 2 t + cos 2 t = 1.
x+1
3
x+1
2
x?1
3
x?1
4
12. Differentiate
a.
d
dx
3xsin 3x + 5
= 1
2
x x + 1 sin 3x + 5
+ 2sin 3x + 5
+ 6x 2 cos 3x + 5
+ 6xcos 3x + 5
x+1
?5x cos x ? 3 + 3cos x ? 3 + 2x 3 sin x ? 3 ? 6xsin x ? 3
= ? 1
2
x
3xsin
3x
+
5
+
sin
3x
+
5
+ 6x 2 cos 3x + 5 ? 6cos 3x + 5
sin 3x + 5 = 1
2
x+1
2
b.
d
dx
c.
d
dx
x2 ? 3
x cos x ? 3
x x + 1 1 ? x ?1
d.
4
d x2 ? 3
dx
x3 + 1
x cos x ? 3
+ 21x 3 cos x ? 3
= ? 1 Ý?11x 5 cos x ? 3
2 x
+ 3cos x ? 3
+ 2x 3 sin x ? 3
? 5x 2 cos x ? 3
+ 2x 6 sin x ? 3
? 6x 4 sin x ? 3
? 6xsin x ? 3 Þ
.
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