1.7 Intermediate Value Theoremap Calculus

Posted By admin On 28/12/21

The idea behind the Intermediate Value Theorem is this:

Therefore, the Intermediate Value Theorem guarantees at least one value c between 0 and 1 with the property that f(c) = 0. This value, of course, is c=ln2. Suggested Reading. Hughes-Hallett et al. Calculus: Single Variable. New York, NY: Wiley. Larson & Edwards. Calculus of a Single Variable: Early Transcendental. The Intermediate Value Theorem (IVT) Formal Statement: If a function is continuous on a closed interval and ( ) ( ), then for every value of between ( ) (and ( ), there exist at least one value of in the open interval ) so that ( ). Translation: A continuous function takes on all the values between any two of its values. Section 4-7: The Mean Value Theorem. In this section we want to take a look at the Mean Value Theorem. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. Show that 5/(x-1) + 7/(x-2) + 16/(x-3) = 0 has root between 1 &2 Please refer the attachment to find what I have done. Please see if it is acceptable.

When we have two points connected by a continuous curve:

  • one point below the line
  • the other point above the line

... then there will be at least one place where the curve crosses the line!

Well of course we must cross the line to get from A to B!

Now that you know the idea, let's look more closely at the details.

Intermediate

Continuous

The curve must be continuous ... no gaps or jumps in it.

Continuous is a special term with an exact definition in calculus, but here we will use this simplified definition:

we can draw it without lifting our pen from the paper

More Formal

Here is the Intermediate Value Theorem stated more formally:

When:

  • The curve is the function y = f(x),
  • which is continuous on the interval [a, b],
  • and w is a number between f(a) and f(b),

Then ...

... there must be at least one value c within [a, b] such that f(c) = w

In other words the function y = f(x) at some point must be w = f(c)

Notice that:

  • w is between f(a) and f(b), which leads to ...
  • c must be between a and b

At Least One

It also says 'at least one value c', which means we could have more.

Here, for example, are 3 points where f(x)=w:

How Is This Useful?

Whenever we can show that:

  • there is a point above some line
  • and a point below that line, and
  • that the curve is continuous,

we can then safely say 'yes, there is a value somewhere in between that is on the line'.

Example: is there a solution to x5 - 2x3 - 2 = 0 between x=0 and x=2?

Frq

At x=0:

05 - 2 × 03 - 2 = -2

At x=2:

25 - 2 × 23 - 2 = 14

Now we know:

  • at x=0, the curve is below zero
  • at x=2, the curve is above zero

And, being a polynomial, the curve will be continuous,

so somewhere in between the curve must cross through y=0

Yes, there is a solution to x5 - 2x3 - 2 = 0 in the interval [0, 2]

An Interesting Thing!

The Intermediate Value Theorem Can Fix a Wobbly Table

If your table is wobbly because of uneven ground ...

... just rotate the table to fix it!

The ground must be continuous (no steps such as poorly laid tiles).

Why does this work?

We can always have 3 legs on the ground, it is the 4th leg that is the trouble.

Imagine we are rotating the table, and the 4th leg could somehow go into the ground (like sand):

  • at some point it will be above the ground
  • at another point it will be below the ground

So there must be some point where the 4th leg perfectly touches the ground and the table won't wobble.

(The famous Martin Gardner wrote about this in Scientific American. There is also a very complicated proof somewhere).

Another One


At some point during a round-trip you will be
exactly as high as where you started.

(It only works if you don't start at the highest or lowest point.)

The idea is:

  • at some point you will be higher than where you started
  • at another point you will be lower than where you started

So there must be a point in between where you are exactly as high as where you started.

Oh, and your path must be continuous, no disappearing and reappearing somewhere else.

The same thing happens with temperature, pressure, and so on.

And There's More!

If you follow a circular path ... somewhere on that circle there will be points that are:

  • directly opposite each other
  • and at the same height!


two points that are
directly opposite and at same height

Can you think of more examples?

In this section we learn a theoretically important existence theorem called theIntermediate Value Theorem and we investigate some applications.

Intermediate Value Theorem

In this section we discuss an important theorem related to continuous functions.Before we present the theorem, lets consider two real life situations and observe animportant difference in their behavior. First, consider the ambient temperature andsecond, consider the amount of money in a bank account.

First, suppose that the temperature is at 8am and then suppose it is at noon.Because of the continuous nature of temperature variation, we can be surethat at some time between 8am and noon the temperature was exactly .Can we make a similar claim about money in a bank account? Supposethe account has $65 in it at 8am and then it has $75 in it at noon. Did ithave exactly $70 in it at some time between 8 am and noon? We cannotanswer that question with any certainty from the given information. On onehand, it is possible that a $10 deposit was made at 11am and so the total inthe bank would have jumped from $65 dollars to $75 without ever beingexactly $70. On the other hand, it is possible that the $10 was added in $5increments. In this case, the account did have exactly $70 in it at some time. Thefundamental reason why we can make certain conclusions in the first casebut cannot in the second, is that temperature varies continuously, whereasmoney in a bank account does not (it will have jump discontinuities). When aquantity is known to vary continuously, then if the quantity is observed to havedifferent values at different times then we can conclude that the quantitytook on any given value between these two at some time between our twoobservations. Mathematically, this property is stated in the Intermediate ValueTheorem.

Intermediate Value Theorem

If the function is continuous on the closed interval and is a number between and ,then the equation has a solution in the open interval .

The value in the theorem is called an intermediate value for the function on theinterval . Note that if a function is not continuous on an interval, then the equation may or may not have a solution on the interval.

Remark: saying that has a solution in is equivalent to saying that there exists anumber between and such that .

The following figure illustrates the IVT.

example 1 Show that the equation has a solution between and .
First, the function is continuous on the interval since is a polynomial. Second,observe that and so that 10 is an intermediate value, i.e., Now we can apply theIntermediate Value Theorem to conclude that the equation has a least one solutionbetween and . In this example, the number 10 is playing the role of in the statementof the theorem.
(problem 1) Determine whether the IVT can be used to show that the equation has asolution in the open interval .

Is continuous on the closed interval ?

and

Is an intermediate value?

YesNo

Can we apply the IVT to conclude that the equation has a solution in the openinterval ?

example 2 Show that the equation has a solution between and .

First, note that the function is continuous on the interval and hence it iscontinuous on the sub-interval, . Next, observe that and so that 2 is anintermediate value, i.e., Finally, by the Intermediate Value Theorem wecan conclude that the equation has a solution on the open interval . Inthis example, the number 2 is playing the role of in the statement of thetheorem.

(problem 2) Determine whether the IVT can be used to show that the equation has a solution in the open interval Is continuous on the closed interval ?

and

Is an intermendiate value?

1.7 Intermediate Value Theoremap Calculus Calculator

YesNo

Does the IVT imply that the equation has a solution in the open interval ?

example 3 Show that the function has a root in the open interval .

Recall that a root occurs when . Since is a polynomial, it is continuous on theinterval . Plugging in the endpoints shows that 0 is an intermediate value: and so By the IVT, we can conclude that the equation has a solution (and hence has aroot) on the open interval .

(problem 3) Determine whether the IVT can be used to show that the function has aroot in the open interval .

Is continuous on the closed interval ?

and

1.7 Intermediate Value Theorem Ap Calculus Calculator

Is an intermendiate value?

YesNo

Does the IVT imply that the function has a root in the open interval ?

example 4 Show that the equation has a solution between and .
Note that the equation is equivalent to the equation . The latter is the prefered formfor using the IVT. So let Since is the difference between two continuous functions, itis continuous on the closed interval . Next, we compute and and show that 0 is anintermediate value: and and so, By the IVT, the equation has a solution in theopen interval . Hence the equivalent equation has a solution on the sameinterval.
1.7 Intermediate Value Theoremap Calculus

Extreme Value Theorem

example 5 Use the IVT four times to approximate a root of the polynomial First, isa polynomial, so it is continuous on any closed interval. Next, note that By the IVT has a root in the interval . To use the IVT a second time, we now determine themidpoint of this interval: and we plug this in to : Combining this with we can usethe IVT again to conclude that has a root on the interval . This intervalis half of the original interval- the original interval has been bisected. Wenow use the IVT a third time. The midpoint of the interval is . We plugthis into : Combining this with we can use the IVT to conclude that hasa root on the interval . This interval is half of the previous interval- theprevious interval has been bisected. We will do this a fourth and final time.Note that is the midpoint of the interval and Combining this with wecan use the IVT a fourth time to conclude that has a root on the interval. At this stage, our approximation of the root is which is the midpointof this interval. Our error is then no more than , which is half the widthof the interval. We will stop here, but the method could theoretically becontinued indefinitely giving a better approximation to the root each time.This method of approximating roots is called the Method of ContinuedBisection.

1.7 Intermediate Value Theorem Ap Calculus Frq

Here is a detailed, lecture style video on the Intermediate Value Theorem: