C ONTINUITY (PA RT III) THE INTE R M E DIATE VALU E THEO R E M In this tutorial, we take a look at a very important property of every typical continuous function. This theorem is called the Intermediate Value Theorem. Did you know that the working principle of a graphing calculator is based on the Intermediate Value Theorem? Well, it is , and later on, we'll briefly see how a typical graphing calculator makes use of this Theorem. You can read more about the Theorem at Wikipedia. You can also view Java Applets illustrating the Theorem at calculusapplets.com. This Theorem is relatively straightforward, and since there isn't any easier way to define it, here's a definition taken directly from the textbook:
Suppose f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b). Then there exists a number c in (a, b) such that f(c) = N.
Note that this Theorem applies to continuous functions only. The graph below illustrates the Theorem:
A GEOMETRIC INTERPR ETATION OF THE INTERMEDIATE VALUE THEOREM If you read the Intermediate Value Theorem carefully, you'll find that it is quite easy to believe. Suppose we have a horizontal line y = N
between f(a) and f(b) [on the next page]. Based on the principle of the
Intermediate Value Theorem, we say that, since f is continuous, then the line y = N WILL DEFINITELY touch f somewhere, and that point of intersection will be (c, f(c)), or (c, N).
The line y = N intersects f here. The same occurs for EVERY continuous function
Next, we look at an application of the Intermediate Value Theorem: locating roots of equations.
APPLICATION S OF THE INTERMEDIATE VALUE THEOREM: LOCATING ROOTS OF EQUATION S To illustrate how the Intermediate Value Theorem can be used to find the roots of an equation, we begin with a simple quadratic equation: f(x) = x2 + 5x + 6. Suppose we want to show that there is a root of f between –2.1 and –1.8. How do we go about it? First, recall that the root of any equation is a number a such that f(a) = 0. On a graph, the root of a equation is the x-coordinate of the point that crosses the x-axis. So, in other words, given the equation x2 + 5x + 6 = 0, we want to prove that a root exists between –2.1 and –1.8.
S OLUTION If you read the problem carefully, it is clear that we are looking for a particular number c between –2.1 and –1.8 such that f(c) = 0. Understand that the problem doesn't exactly require that we find the value of c, but instead, we prove that such number exists within the specified interval [–2.1, –1.8]. Using the Theorem, we find that a = –2.1
b = –1.8
f(x) = x2 + 5x + 6
Remember that this Theorem works only for continuous functions. Therefore, we need to confirm that the function we're dealing with is continuous. Since f is a polynomial, then we know by the continuity theorems that it is continuous everywhere. The next thing we have to do is obtain the values of the endpoints of the interval, that is, f(a) and f(b), then we compare these values with reference to N (that is, whether the values of f(a) and f(b) are less than or greater than N, as the situation dictates): f(a) = f(–2.1) = (–2.1)2 + 5(–2.1) + 6 = –0.09
(which is less than N)
f(b) = f(–1.8) = (–1.8)2 + 5(–1.8) + 6 = 0.24
(which is greater than N)
The calculations above show that f(–2.1) < N < f(–1.8). At this point, it is very clear that the number we are looking for does exist between –2.1 and –1.8. Alternatively, we can say that the equation x 2 + 5x + 6 = 0 has at least one real root in the interval [–2.1, –1.8]. By applying the Intermediate Value Theorem again, we can locate c more precisely. This time, we use even closer (but not too close) values. For example, f(–2.0001) = (–2.0001)2 + 5(–2.0001) + 6 = –0.0009999 < N f(–1.9999) = (–1.9999)2 + 5(–1.9999) + 6 = 0.00010001 > N Using the Theorem, we see that a root exists in the interval (–2.0001, –1.9999). The equation x2 + 5x + 6 = 0 actually has two roots –2 and –3. So, in the example above, the number c we were looking for was –2, although we were not required to find it!! Using a graphing device can also help you confirm your results. The figure below shows the graph of f in the viewing rectangle [–2.5, 0.6] by [–0.6, 1]:
By zooming in, you are more likely to locate the root. Zooming in to the viewing rectangle [–2.2, –1.7] by [–0.2, 0.5] gives this:
I hope you understand the basic concept of this Theorem and how to use it. We'll be treating several examples later in this tutorial. But before we continue, here's a very brief description of how graphing calculators use the Intermediate Value Theorem do perform their graphing functions: The computer calculates a finite number of points on the graph, It then turns on the pixels that contain these calculated points. The computer finally connects the pixels by turning on the intermediate pixels. This, of course, is based on the assumption that the function is continuous and takes on all intermediate values between two consecutive points.
EXAMPLE 2 Use the Intermediate Value Theorem to show that there is a root of the equation x3 – 3x + 1 = 0 between 0 and 1.
S OLUTION The first step is to identify the given values. Let g(x) = x3 – 3x + 1 = 0 a = 0, b = 1 and N = 0. Based on the Intermediate Value Theorem, the function g has at least one real root in the interval (0, 1). We confirm this by computing g(a) and g(b): g(0) = (0)3 – 3(0) + 1 = 1 > 0 g(1) = (1)3 – 3(1) + 1 = –1 < 0 Thus, we see that g(1) < N < g(0) OR
g(1) < 0 < g(0)
This confirms that the number we're looking for does exist in the interval (0, 1). To locate c more precisely, we use a smaller interval. So, let's put a = 0.3 and b = 0.7. Thus, the interval is (0.3, 0.7), and we find that g(0.3) = (0.3)3 – 3(0.3) + 1 = 0.127 > 0 g(0.7) = (0.7)3 – 3(0.7) + 1 = –0.757 < 0 Again, we find that g(0.7) < 0 < g(0.3). This confirms that the root lies in the interval (0.3, 0.7). Let's push our luck and use an even smaller interval: (a, b) = (0.35, 0.4), so that g(0.35) = (0.35)3 – 3(0.35) + 1 = –0.007125 < 0 g(0.4) = (0.4)3 – 3(0.4) + 1 = –0.136 < 0 What does the last calculation tell you? Perhaps we have pushed a little too far. The last calculation shows that f(a) and f(b) are less that zero, which means that the root is DEFINITELY NOT within the interval (0.35, 0.4). This should tell you that you need to be absolutely careful when choosing intervals. Let's try one more interval. This time, we'll make the interval relatively larger than the previous one. Let's say (a, b) = (0.32, 0.37) so that g(0.32) = (0.32)3 – 3(0.32) + 1 = 0.072768 > 0 g(0.37) = (0.37)3 – 3(0.37) + 1 = –0.059347 < 0 We see that g(0.37) < 0 < g(0.32), and this implies that a root lies in the interval (0.32, 0.37). This will do for now. Of
course, we could always use a slightly smaller interval. Like I mentioned earlier, DON”T PUSH IT!! A graphing calculator can help you confirm your results and even help you determine what intervals to use. Graphing g in the viewing rectangle [–3, 3] by [–10, 10] produces this:
As you can see, a root lies between 0 and 1. Zooming in to the viewing rectangle [0.3, 0.43] by [–0.2, 0.2] produces this:
This proves that our calculations were correct. Using a calculator, we see that the root of g is roughly 0.345. Observe
that g actually has three roots, but we're concerned about the root that lies between 0 and 1.
EXAMPLE 3 Use the Intermediate Value Theorem to show that there is a root of the equation x3 = √x + 1 between 1 and 2.
S OLUTION First, identify the given values: f(x) = x3 = √x + 1 a = 1, b = 2, N = ? We are looking for a solution of f, that is, a number c such that f(c) = 0. Thus, we need to rewrite f To be able to determine the root(s) of f , we need to express f in such a way that f(x) = 0. Thus, x 3 = √x + 1 becomes x 2 – √x + 1 = 0 Now, we know that N = 0. Next, we determine the values of f(a) and f(b). f(a) = f(1) = (1)2 – √1 + 1 = 1 – √2 = –0.4142 < 0 f(b) = f(2) = (2)2 – √2 + 1 = 4 – √3 = 2.2679 > 0 Since f(1) < 0 < f(2), we deduce that f definitely has a root in the interval (1, 2). So, the next thing to do is use a smaller interval, but how do we know what interval(s) to choose? HINT: Use a calculator to compute the exact value of the root and then use values that are slightly less and greater than the root to prove that the Intermediate Value Theorem works for the equation you're dealing with. For example, if the root of a certain equation g is 1.3, then choose your interval (a, b) to be something like (1.25, 1.31). You get the idea!! So, let's use another interval (1.21, 1.23), which means a = 1.21 and b = 1.23. Therefore, f(a) = f(1.21) = (1.21)2 – √1.21 + 1 = –0.00225 < 0 f(b) = f(1.23) = (1.23)2 – √1.23 + 1 = 0.0196 > 0 We see that f(1.21) < 0 < f(1.23), which means that a root lies in the interval (1.22, 1.23). By trial and error, f(1.22) = –0.001566 < 0 f(1.221) = 0.0000539 > 0 Thus, a root surely lies in the interval (1.22, 1.221).
EXAMPLE 4 Use the Intermediate Value Theorem to show that there is a root of the equation cos x = x between 0 and 1.
S OLUTION We see that F(x) = cos x = x a = 0, b = 1, N = ? We don't know N because has not been properly expressed. Since we are to look for a root of f , we should express f to equal zero. In other words, we rewrite to to look like this: cos x – x = 0 Which means N = 0. Next, we compute F(a) and F(b):
F(a) = F(0) = cos 0 – 0 = 1 – 0 = 1 > 0 F(b) = F(1) = cos 1 – 0 = 0.5401 – 1 = –0.4597 Since F(1) < N < F(0), then a root lines between 0 and 1. A calculator gives, by trial and error, F(0.73) = 0.01517 > 0 F(0.74) = –0.001531 Since F(0.74) < 0 < F(0.73), a root must surely lie in the interval (0.73, 0.74). A graph can always help you clarify.
EXAMPLE 5 Use the Intermediate Value Theorem to show that there is a root of the equation 2sin x = 3 – 2x between 0 and 1.
S OLUTION First, we rewrite the equation 2sin x – 3 + 2x = 0 OR 2sin x + 2x – 3 = 0 Therefore, G(x) = 2sin x + 2x – 3,
a = 0, b = 1, N = 0.
And, G(0) = 2sin 0 + 2(0) – 3 = 0 + 0 – 3 = – 3 < 0 G(1) = 2sin 1 + 2(1) – 3 = 1.68294 + 2 – 3 = 0.68294 > 0
Since G(0) < 0 < G(1), then the function G must have a root between 0 and 1. Now, let's use a smaller interval (0.78, 0.79). Using a calculator, we see that G(0.78) = – 0.03344 < 0 G(0.79) = 0.0007065 > 0 Again, we see that G(0.78) < 0 < G(0.79), which implies that a root lies in the interval (0.78, 0.79), as the graph below illustrates:
Graph of G(x) = 2sin x + 2x – 3 in the viewing rectangle [-10, 10] by [-10, 10]
Now, look at this:
Graph of G(x) = 2sin x + 2x – 3 in the viewing rectangle [0.75, 0.8] by [-0.2, 0.2]
EXAMPLE 6 If sin x = 2 – x, (a).
Prove that the equation has at least one real root
Use your calculator to find an interval of length 0.01 that contains a root.
S OLUTION First, we have to rewrite the function in such a way that it equals zero. This way, we'll be able to find the root. So, we rewrite sin x = 2 – x as sin x – 2 + x = 0. Therefore, f(x) = sin x – 2 + x, and N = 0, but we are not sure of what interval to use. What we need to do now is find the exact value of the root of f. Using a calculator, we find that the root of f is approximately 1.1061, as the graph illustrates:
Okay, now that we know the root of f, we can safely choose an interval, in which we'll prove that a root exists. Let's pick an interval like, say, (1.1, 1.12). This means a = 1.1 and b = 1.12 So that f(1.1) = - 0.0087926 < 0 f(1.12) = 0.0201004 > 0 At this point, we see that f(1.1) < 0 < f(1.12), which confirms that a root exists in the interval (1.1, 1.12). The second task requires that we choose an interval of length 0.01 that contains a root. In other words, we should choose an interval (a, b) such that the difference between the endpoints a and b is 0.01. The root of f is given as 1.1061 (correct to 4 decimal places). Again, remember that the difference between the endpoints a and b is 0.01, which means that a and b must be correct to 2 decimal places. Now the root is 1.1061. therefore, a < 1.1061 < b If we correct 1.1061 to 2 decimal places, we have 1.11. Obviously, this would be the endpoint b, since it is greater than the root. Therefore, the value of endpoint a would be 1.11 – 0.01 = 1.10 Hence, the interval we are looking for is (1.10, 1.11), and you can see that the root fits nicely into this interval.
Graph of sin x = 2 – x in the viewing rectangle [1.09, 1.13] by [-0.05, 0,05]
EXAMPLE 7 Use the Intermediate Value Theorem to prove that there is a positive number c such that c2 = 2.
S OLUTION We are required to prove that there is a positive number c such that c2 = 2. To do this, we write the equation in such a way that allows us to use the Intermediate Value Theorem. Thus we have
c2 – 2 = 0 When we solve this equation, we find that there are two possible values for c: – √2 and √2. Below is graphical proof:
Graph of f(x) = c2 – 2 in the viewing rectangle [-3, 3] by [-10, 10]. The dotted points on the curve represent the roots of f: – √2 and √2
Notice that that the curve intersects the x-axis at two points, as represented by the two red dots on the graph. These points mark the roots of f. notice that we are asked to prove that a positive number c exists such that c2 – 2 = 0. This means that we have to deal with only the positive root: √2. At this point, we now know for a fact that c = √2, but how do use the Intermediate Value Theorem to prove that it exists? Simple!!! We start with the graph. Notice that the root lies between 1 and 2, so we use the interval (1, 2) for starters. Therefore, a = 1, b = 2,
which gives f(a) = f(1) = 12 – 2 = –1 < 0 f(b) = f(2) = 22 – 2 = 2 > 0 This confirms that c lies in the interval (1, 2). Let's try a “tighter” interval: (1.3, 1.5). f(a) = f(1.3) = (1.3)2 – 2 = –0.31 < 0 f(b) = f(1.5) = (1.5)2 – 2 = 0.25 > 0 This shows that c lies in the interval (1.3, 1.5). Let's try another interval: (1.35, 1.45). f(a) = f(1.35) = (1.35)2 – 2 = –0.1775 < 0 f(b) = f(1.45) = (1.45)2 – 2 = 0.1025 > 0 We see that f(1.35) < 0 < f(1.45), and therefore the Intermediate Value Theorem says that there is a number c between 1.35 and 1.45 such that f(c) = 0. What all this means is that the function c 2 =2 has one positive root in the interval (1.35, 1.45) – which we already know is √2,
EXAMPLE 8 If f(x) = x3 – x2 + x, show that there is a number c such that f(c) = 10.
S OLUTION If you read the question carefully, you'll see that we are asked to show that x3 – x2 + x = 10 or more precisely, we want to show that there is a number c such that x3 – x2 + x – 10 = 0 where f(x) = x3 – x2 + x – 10 and c is the root of f. We start by graphing f:
Graph of f in the viewing rectangle [1.95, 3] by [-10, 10]. From the graph, it is clear that that a root lies between 2.3 and 2.4, and so our interval of interest will be (2.3, 2.4). Therefore, a = 2.3, b = 2.4, and N = 0, so that f(a) = f(2.3) = (2.3)3 – (2.3)2 + 2.3 – 10 = –0.823 < 0 f(b) = f(2.4) = (2.4)3 – (2.4)2 + 2.4 – 10 =
0.464 > 0
Observe that f(2.3) < 0 < f(2.4), which confirms the existence of a root between 2.3 and 2.4. In other words, for the function f(x) = x3 – x2 + x – 10, there is a number c that lies between 2.3 and 2.4 such that f(c) = 10.
EXAMPLE 9 Is there a number that is exactly one more than its cube?
S OLUTION Interesting question!!!!! When faced with a problem like this. the key to working it out is to write the word problem in a form you can easily understand. Let's take a look at the statement, and translate it piece by piece: “ a number that is exactly one more than its cube” Two numbers are involved here: a number (let's call it c), and the cube of the number (which would be c3). These two numbers differ by 1. So, the equation we're looking for is c + 1 = c3 OR
c = c3 – 1
Better yet, we can write it as c3 – c – 1 = 0 Now the question becomes: what is/are the root(s) of the equation?
Let's start by graphing it (let's assume f(c) = c3 – c – 1):
Looks like our root lies between 1 and 1.5. Let's confirm: f(1) = –1 < 0 f(1.5) = 0.8750 > 0 This confirms that the number we're looking for exists between 1 and 1.5. Let's zoom in and see what we find:
This time we are much closer to our target. Remember that we are only to show that the number exists, We don't need to know what it is. The graph shows that the root lies between 1.3245 and 1.325. Let's verify: f(1.3245) = –0.0009293 < 0 f(1.325)
0.0020313 > 0
Whatever the number c is, at least we know it exists, and lies between 1.3245 and 1.325.
EXAMPLE 10 A Tibetan Monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.
S OLUTION To solve this â€œmysteryâ€? we need to visualize the Monk's motion as he moves back and forth on the mountain. At this point, you should understand exactly what the question is asking you to do. Let's dissolve the problem: the Monk leaves the monastery at 7:00 AM and gets to the top of the mountain at 7:00 PM. He probably stays there for the next twelve hours, and then starts back home the following morning at 7:00 AM (taking the very same path he took the previous day), arriving the monastery at 7:00 PM. Now, the question is this: what point on that path does the Monk cross at PRECISELY the same time of day on both days? Sounds like a graph would be the best tool the solve this problem:
Mon k stays on moun tain top for 12 h ours Mon k gets to top of moun tain At 7 PM
Mon k leaves Mon astery At 7 AM
Mon k arrives at Mon astery At 7 PM
Mon k goes to Top of moun tain
Mon k goes back to Mon astery
Mon k leaves Moun tain top At 7 AM
The following is assumed from the graph above: ➔ The path is a straight one ➔ The x-axis represents the ground level (which is where the monastery would be) ➔ The Monk is moving on the mountain a relatively consistent rate (on both days) ➔ The mountain is 100 feet high. The numbering on the x-axis looks weird, but when you compare it with the question, you'll see it makes perfect sense. Using the graph let's analyze the Monk's motion on a hourly basis: Day 1: 7:00 AM – He leaves the monastery 8:00 AM – about 9 feet up the mountain 9:00 AM – about 15 feet up the mountain 10:00 AM – about 23 feet up the mountain 11:00 AM – about 32 feet up the mountain 12:00 PM – about 40 feet up the mountain 1:00 PM – about 50 feet up the mountain 2:00 PM – about 58 feet up the mountain 3:00 PM – about 66 feet up the mountain 4:00 PM – about 74 feet up the mountain 5:00 PM – about 82 feet up the mountain 6:00 PM – about 90 feet up the mountain 7:00 PM – 100 feet (top of mountain) Day 2: 7:00 AM – He leaves the mountain top 8:00 AM – about 10 feet down the mountain 9:00 AM – about 18 feet down the mountain 10:00 AM – about 24 feet down the mountain 11:00 AM – about 33 feet down the mountain 12:00 PM – about 42 feet down the mountain 1:00 PM – about 50 feet down the mountain 2:00 PM – about 60 feet down the mountain 3:00 PM – about 68 feet down the mountain 4:00 PM – about 77 feet down the mountain 5:00 PM – about 84 feet down the mountain 6:00 PM – about 92 feet down the mountain 7:00 PM – Back at the Monastery (ground level)
Let's assume the upward motion of the Monk is represented by the function y = f(x) and the downward motion by y = g(x). Then the intermediate Value Theorem says that there is a number c in f and g such that f(c) = g(c). From the analysis above, we see that the Monk crossed the 50ft mark at exactly 1.00 PM on both days. In the context of the Intermediate Value Theorem, we say that f(1) = g(1) = 50. As you can see on the graph, x marks the spot.
EXE R CI SE S (1) Suppose that a function f is continuous on [0, 1] except at 0.25 and that f(0) = 1 and f(1) = 3. Let N = 2. Sketch two possible graphs of f, one showing that f might not satisfy the conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn't satisfy the hypothesis). (2) Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. (I). tan x = 2x
(II). 2x + x + 2 = 0
(3) Prove that the equation has a least one real root. Use your graphing device to find the root correct to three decimal places. (I).
x5 – x4 – 4 = 0
√x – 5
And with that, we conclude this section. Next, we look at Tangents, Velocities and Rates of Change.