Download The Intermediate Value Theorem DEFINITIONS Intermediate means

The Intermediate Value Theorem
DEFINITIONS
Intermediate means “in-between”. In this case, intermediate means between two
known y-values. The word value refers to “y” values.
A theorem: “…is a statement that can be demonstrated to be true by accepted
mathematical operations and arguments”1 .
So the Intermediate Value Theorem is a theorem that will be dealing with all of
the y-values between two known y-values.
BRIEFLY STATED
If a function produces one y-value larger than another one, then it will produce
all of the y-values between these two as long as the function has no holes, rips, or
tears in it. In other words, it is guaranteed that there will be x-values that will
produce the y-values between the other two if the function is continuous.
Example 1:
2
Let’s use y = x +1 for x-values between 1 and 5 as an example.
2
First, y = x +1 is a smooth curve that has no rips, tears, or holes in it, so we call it
continuous.
2
2
If you put x = 1 into x +1, it will produce y = 2. And if you use x = 5, then x +1
will produce y = 26.
2
Thus the Intermediate Value Theorem will guarantee that the function y = x +1 will
produce all of the real numbers between 2 and 26. Furthermore, the Intermediate
Value Theorem guarantees that these y-values will be produced by numbers chosen
for x between 1 and 5.
Remember, real numbers are ALL numbers such as fractions, decimals, integers,
irrational numbers, square roots, etc. They are not imaginary numbers like a+bi.
So what’s the BIG deal?
1
Source: Weisstein, Eric W. "Theorem." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/Theorem.html
The BIG deal is how to mathematically justify the Intermediate Value Theorem.
So we study the mathematics behind the theorem.
INTERMEDIATE VALUE THEOREM2
“If f is continuous on a closed interval [a, b], and c is any number between f(a) and
f(b), then there is at least one number x in the closed interval such that f(x) = c”.
BACKGROUND
“If f is continuous … and c is any number between f(a) and f(b)…”
What does this mean?
When we have a function, we use “x” values to produce “y” values.
Usually we focus on “x” (the domain) first because we put “x” into the function, do the
arithmetic, and get the y-value. We name the function f, or g, or h, or whatever.
When we put a number, say 6, into the function, we write f(6) and this produces a yvalue.
For example,
If x = 5 and y = f ( x) = 3 x 2 + 1 then, y = 3(5)2+1 = 3(25)+1 = 75 +1 = 76
So we would write: f(5) = 76. This means the value of y will be 76 when x is 5.
In the theorem, f(a) and f(b) are simply y-values.
f(a) is the y-value when x = a.
f(b) is the y-value when x = b.
The statement “c is any number between f(a) and f(b)” means that c is also a y-value but
it is between the other two. This is a little confusing because it is written without f( ).
Symbolically, f (a ) < c < f (b)
2
or
f (b) < c < f (a )
Renze, John and Weisstein, Eric W. "Intermediate Value Theorem." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/IntermediateValueTheorem.html
With the Intermediate Value Theorem, we are working with a function f.
The theorem encourages us to begin by looking at y-values and thinking about (but not
finding) the x-values that produced them.
Example 2:
Let’s work with: y = f ( x) = 3 x 2 + 1
It too produces a smooth graph with no holes, tears, or rips. It too is continuous.
Just for fun, let’s look at two y-values. Let’s look at y = 76 and y = 13.
In this case, if you knew y = 76 then there had to be an “x” that produced it.
If you wanted to find “x”, all you’d have to do would be to set the function 3x2+1 equal
to 76 and solve for x ( in this case, using quadratic methods).
The point of the Intermediate Value Theorem is that you are guaranteed a y-value
between two others. This will occur at some x-value, and it doesn’t matter that we don’t
know the “x”.
So we could write f(b) = 76. The unknown x-value would be the “b”.
If you also knew y = 13. then we would know that there was another different x that
produced it; in other words f(a) = 13 where “a” is the other x-value.
Since you knew that the function produced y = 76 and it also produced y = 13, it would
be pretty obvious that the function will produce all of the y-values between 13 and 76.
In this context, the statement “c is any number between f(a) and f(b)” means “c” is any
y-value between 13 and 76.
HYPOTHESES
In short, the last example illustrates the hypotheses, or assumptions, of the Intermediate
Value Theorem.
We know that the function is continuous.
We know that the function produces one y-value greater than another one.
We also know that the function will produce ALL of the y-values between them, so there
will be a y-value between the other two.
CONCLUSION or RESULT
Thus we could conclude that there will be an x-value, that will produce some y-value
between the other two.
USING THE THEOREM.
As we said earlier, the BIG deal is how to mathematically justify the Intermediate Value
Theorem.
You will need to show two things to verify the hypothesis:
1. That the function f(x) is continuous on an interval [a, b]
2. That one value of y is greater than a second value of y when these values are
produced by x from the interval [a, b] AND that a third y-value lies between the
other two.
Once you confirm both of these, then you can say: “there is at least one number x in the
closed interval such that f(x) = c”. This is the conclusion of the theorem.
INTERMEDIATE VALUE THEOREM
“If f is continuous on a closed interval [a, b], and c is any number between f(a) and
f(b), then there is at least one number x in the closed interval such that f(x) = c”.
Example 3:
Continuing the earlier example, y = f ( x) = 3 x 2 + 1 along with y = 76 and y = 13
Will this function ever equal 27.279? In other words, will y = 27.279?
Step 1. Is the function continuous on [a, b]?
It is a continuous function because it is a polynomial function and all polynomial
functions are continuous for all real numbers. Since [a, b] is a subset of all real numbers
on the x-axis, then the function is also continuous on [a, b].
To show that a function is continuous, you must be sure there are no holes, rips, or tears
in the function. For the calculus students, if a group of functions has not been designated
as continuous, then you will have to “prove” continuity by showing for all points “x = k”
that:
•
•
f(k) exists
lim f ( x) exists
•
lim f ( x) =f(k)
x →k
x →k
As we said earlier, if a function is a polynomial function, this group of functions has
already been shown in the mathematical community to be continuous for all real
numbers.
So the function y = f ( x) = 3 x 2 + 1 is continuous on [a, b].
But what are “a” and “b” ?
They are the left and right end-points of a closed interval on the x-axis.
How do we find their specific values?
By inspection, x = 5 produces a y-value of 76.
In other words, 3(5)2+1 = 3(25) + 1 = 75 + 1 = 76
We can solve 3x2+1 = 13 to get x = ± 2. So x = 2 produces a y-value of 13.
Thus we could define an interval for the function as [2, 5] ;
“a” would be 2 and “b” would be 5.
Hence, the function y = f ( x) = 3 x 2 + 1 is continuous on [2, 5].
Step 2. Next, is one value of y greater than a second value of y when these values are
produced by x from the interval [2,5] ?
It is clear that a y-value of 76 is larger than a y-value of 13.
Notice that 76 was produced by x = 5 and 13 was produced by x = 2.
Both of these x-values are from the endpoints of [2. 5].
So the answer is “yes”, one value of y is greater than a second value of y on [2. 5].
Step 3. Finally, is the third y-value, 27.279, between the other two?
Yes.
13 < 27.279 < 76
Using the Intermediate Value Theorem:
•
•
the function y = f ( x) = 3 x 2 + 1 is continuous on [2, 5].
f(2) = 13 < 27.279 < 76 = f(5).
Thus: “there is at least one number x in the closed interval [2, 5] such that
y = f ( x) = 3 x 2 + 1 = 27.279”. And we’re done.
INTERMEDIATE VALUE THEOREM
“If f is continuous on a closed interval [a, b], and c is any number between f(a) and
f(b), then there is at least one number x in the closed interval such that f(x) = c”.
Example 4
This time let’s use: y = f ( x) = x3 − 3 and ask the question will this function ever equal
zero on the interval [-4, 3] ?
Step 1. Is the function continuous on [-4, 3]?
It is a continuous function because y = f ( x) = x3 − 3 is a polynomial function and all
polynomial functions are continuous for all real numbers. Since [-4, 3] is a subset of all
real numbers on the x-axis, then the function is also continuous on [-4, 3].
Step 2. Is one value of y greater than a second value of y when these values are produced
by x from the interval [-4, 3] ?
f (−4) = (−4)3 − 3 = −64 − 3 = −67
f (3) = (3)3 − 3 = 27 − 3 = 24
−67 < 24
One value of y is greater than a second value of y when these values are produced by x
from the interval [-4, 3].
Step 3. Finally, is the y-value that we wanted, y = 0, between the other two?
−67 < 0 < 24
Yes!
Using the Intermediate Value Theorem:
•
•
the function y = f ( x) = x3 − 3 is continuous on [-4, 3].
f (−4) = −67 < 0 < 24 = f (3)
Thus: “there is at least one number x in the closed interval [-4, 3] such that
y = f ( x) = x3 − 3 = 0 ”.
Example 5:
Use the Intermediate Value Theorem to show that f ( x) = x 4 + 5 x3 − 3 x 2 − 8 x
will equal 6 somewhere between x = 0 and x = 5.
Strategy:
We could randomly try x-values between x = 0 and x = 5, but that would be a waste of
time. If we use the Intermediate Value Theorem, all we need to do is to show that there is
a function f(x) that is continuous on [0, 5] and that the number we want is between f(0)
and f(5).
The easiest way to get a number between f(0) and f(5) is to rewrite the function so it
equals 0 and see if f(x) changes signs (from positive to negative or vice versa) on [0, 5].
If one value of the function is positive to one side of y = 0, and another value of the
function is negative on the other side of y = 0, then we’d know for sure that y = 0 was
between the other two. Note: negative < zero < positive.
The way we come up with y = 0 is to rewrite the given function so it equals zero.
In our example, we want to know when f ( x) = x 4 + 5 x3 − 3 x 2 − 8 x = 6 .
If we subtract 6 from both sides of the equation, we create a new function
f1 ( x) = x 4 + 5 x 3 − 3 x 2 − 8 x − 6 = 6 − 6 = 0 which, as you can see, equals zero.
We will apply the conditions for the Intermediate Value Theorem to this new
function f1 ( x) = x 4 + 5 x 3 − 3 x 2 − 8 x − 6 .
If the conditions are met, then there will be an x from [0, 5] that will make the new
function equal 0.
And if the new function equals zero, then we can add 6 to both sides of it in order to
obtain the original function with which we started.
INTERMEDIATE VALUE THEOREM
“If f is continuous on a closed interval [a, b], and c is any number between f(a) and f(b),
then there is at least one number x in the closed interval such that f(x) = c”.
Solution:
Step 1.
f1 ( x) = x 4 + 5 x 3 − 3 x 2 − 8 x − 6 is a polynomial function, so it is continuous on [0, 5].
Step 2.
Let c = 0. In other words, f1 ( x) = x 4 + 5( x)3 − 3( x) 2 − 8( x) − 6 = 0
Step 3.
Is c = 0 between f(0) and f(5)?
f1 (0) = 04 + 5(0)3 − 3(0) 2 − 8(0) − 6 = −6
f1 (5) = 54 + 5(5)3 − 3(5)2 − 8(5) − 6 = 1129
This shows that f1 (0) = −6 < c < 1129 = f1 (5)
Since f is continuous on a closed interval [0, 5], and 0 is a number between f(0) and f(5)
then by the Intermediate Value Theorem there is at least one number x in the closed
interval such that f(x) = c = 0.
This means we have shown that f1 ( x) = x 4 + 5( x)3 − 3( x) 2 − 8( x) − 6 = 0
for some x on [0, 5]
Algebraically, x 4 + 5( x)3 − 3( x) 2 − 8( x) − 6 + 6 = 0 + 6
= x 4 + 5( x)3 − 3( x) 2 − 8( x) = 6
So, the original function
f ( x) = x 4 + 5 x3 − 3 x 2 − 8 x will equal 6 on [0, 5].