WEBVTT
1
00:00:01.240 --> 00:00:03.330 A:middle L:90%
this problem Number 60 of the Stuart crack the safe
2
00:00:03.330 --> 00:00:07.669 A:middle L:90%
edition Section 2.5 a party proved that the equation has
3
00:00:07.669 --> 00:00:09.859 A:middle L:90%
at least one real root in part B. Use
4
00:00:09.859 --> 00:00:12.109 A:middle L:90%
your graphing device to find the route correct to three
5
00:00:12.109 --> 00:00:15.929 A:middle L:90%
decimal places. The equation, given his arc Tangent
6
00:00:15.929 --> 00:00:19.679 A:middle L:90%
of X, is equal to one minus X on
7
00:00:19.679 --> 00:00:23.570 A:middle L:90%
our first step will be to rearrange this equation so
8
00:00:23.570 --> 00:00:26.410 A:middle L:90%
that all the terms on the left side mark 10
9
00:00:26.410 --> 00:00:33.950 A:middle L:90%
X plus X minus one course zero or the left
10
00:00:33.950 --> 00:00:39.130 A:middle L:90%
side we consider is the function F and any roots
11
00:00:39.140 --> 00:00:42.829 A:middle L:90%
are the values of X that make this function equal
12
00:00:42.950 --> 00:00:45.929 A:middle L:90%
to zero for party. We're just going to prove
13
00:00:45.929 --> 00:00:47.570 A:middle L:90%
that these exists in part, we will be able
14
00:00:47.570 --> 00:00:51.259 A:middle L:90%
to find what the root is using our crafting calculators
15
00:00:52.939 --> 00:00:58.109 A:middle L:90%
. The the approach to proving that this function has
16
00:00:58.109 --> 00:01:00.880 A:middle L:90%
at least one route is to understand the behaviour of
17
00:01:00.880 --> 00:01:04.530 A:middle L:90%
this function. For a certain interval, we're going
18
00:01:04.530 --> 00:01:07.189 A:middle L:90%
to choose, Do you, you know, largest
19
00:01:07.189 --> 00:01:14.250 A:middle L:90%
interval possible and understand what this function looks like when
20
00:01:14.250 --> 00:01:18.109 A:middle L:90%
we approach infinity and also negative infinity. So for
21
00:01:18.109 --> 00:01:21.959 A:middle L:90%
the first limit as X approaches infinity, the function
22
00:01:23.040 --> 00:01:26.650 A:middle L:90%
approaches infinity because this function is our tangent, which
23
00:01:26.659 --> 00:01:30.750 A:middle L:90%
only has a limited range range and a maximum value
24
00:01:30.750 --> 00:01:34.950 A:middle L:90%
of power to plus infinity X approaches infinity minus one
25
00:01:36.209 --> 00:01:38.269 A:middle L:90%
. So this whole function will be dominated by this
26
00:01:38.269 --> 00:01:42.590 A:middle L:90%
X term and it will be infinity. Likewise,
27
00:01:42.590 --> 00:01:45.819 A:middle L:90%
as we approach negative infinity, the function will be
28
00:01:45.819 --> 00:01:49.590 A:middle L:90%
dominated by the X term as well, and this
29
00:01:49.590 --> 00:01:56.049 A:middle L:90%
function approaches negative infinity. So the the reason that
30
00:01:56.739 --> 00:01:59.409 A:middle L:90%
we can say this function has at least one route
31
00:01:59.409 --> 00:02:01.040 A:middle L:90%
is because through the intermediate value theorem, because this
32
00:02:01.040 --> 00:02:04.659 A:middle L:90%
function is a continuous function. Since the combination of
33
00:02:05.040 --> 00:02:07.909 A:middle L:90%
trigonometry functions, linear functions and constant values, it
34
00:02:07.909 --> 00:02:12.379 A:middle L:90%
is a continuous function, and for this given interval
35
00:02:12.379 --> 00:02:14.870 A:middle L:90%
from infinity to infinity, it must take on every
36
00:02:14.870 --> 00:02:17.509 A:middle L:90%
single value of what the function is evaluated at.
37
00:02:17.520 --> 00:02:21.159 A:middle L:90%
So it's negative. Infinity and negative infinity. It's
38
00:02:21.159 --> 00:02:23.939 A:middle L:90%
positive affinity, a positive infinity. Since it's continuous
39
00:02:23.939 --> 00:02:24.650 A:middle L:90%
, we know that they must take on every single
40
00:02:24.650 --> 00:02:29.379 A:middle L:90%
value between these two numbers, including zero, so
41
00:02:29.379 --> 00:02:30.770 A:middle L:90%
there must be at least one real root. According
42
00:02:30.770 --> 00:02:35.560 A:middle L:90%
to the Intermediate Value Theorem, we will graph in
43
00:02:35.560 --> 00:02:38.150 A:middle L:90%
part be using your graphing calculator. You can either
44
00:02:38.639 --> 00:02:42.199 A:middle L:90%
plot this function equal to this function and see where
45
00:02:42.199 --> 00:02:44.830 A:middle L:90%
they intersect. Or just take this function that we
46
00:02:44.840 --> 00:02:47.120 A:middle L:90%
called f plot it and see where it crosses the
47
00:02:47.120 --> 00:02:51.120 A:middle L:90%
X axis because that's where X or why why will
48
00:02:51.120 --> 00:02:53.210 A:middle L:90%
be equal to zero? And we can figure out
49
00:02:53.210 --> 00:02:55.259 A:middle L:90%
exactly what the value is to three decimal places.
50
00:02:55.840 --> 00:03:00.030 A:middle L:90%
We see that the function crosses the X axis at
51
00:03:00.030 --> 00:03:09.039 A:middle L:90%
0.5 to 3, or approximately 0.52 Oh, is
52
00:03:09.039 --> 00:03:14.060 A:middle L:90%
what we will consider the solution to this problem.
53
00:03:14.740 --> 00:03:17.740 A:middle L:90%
This is a corrected through decimal places where the route
54
00:03:17.750 --> 00:03:21.599 A:middle L:90%
occurs, where the function will be equal to zero
55
00:03:21.789 --> 00:03:23.400 A:middle L:90%
, and we can definitely double check that I plug
56
00:03:23.400 --> 00:03:23.939 A:middle L:90%
them into the equation, and we should get a
57
00:03:23.939 --> 00:03:25.259 A:middle L:90%
value very close to zero.