### Inside the IVT

#### What the IVT Does and Does Not Say.

The IVT is a powerful tool, but it has its limitations. To illustrate, suppose that d(t) represents the decibel level of Pork Chop's motorcycle engine, and suppose that you know that

• d(0) = 100 and d(10) = 35, where t is measured in seconds.

• d is a continuous function.
The IVT allows us to conclude that in the ten second interval between time t=0 and time t=35 Pork Chop's decibel level reached every value between 35 and 100. But it does NOT say anything about

• When or how many times (other than at least once) a particular decibel was attained.

• Whether or not decibel levels bigger than 100 or less than 35 were reached.

The following graphs of PC's noise illustrate that very different behaviors are consistent with the hypothesis that d(t) is continuous and that its values at 0 and 10 are 100 and 35 respectively.

#### Why Graphs Cross

In analyzing your trip up the knoll, your guide reached the conclusion that the graphs representing your trip up and trip down had to cross. This used the continuity of the two functions being graphed and is a consequence of the IVT:

 If f(x) and g(x) are two functions which are continuous on the closed interval [a,b], and if f(a) < g(a) and f(b) > g(b), then there is some c in [a,b] where f(c) = g(c).

Why is THAT true?

#### Why is continuity necessary?

The function j(t) encountered in Stage 3 and graphed at the right has a jump discontinuity at t=-4. The IVT fails for this function! Let a = -5 and b = -3. Then j(-5) is about .5 and j(-3) is about -.5, but there is no t0 between -5 and -3 for which j(t0) = .25.

Or think of what might have happened if you could have called to your spaceship when it started to rain on the knoll. "Beam me to the bottom!" you would have said, and in the next instant you would have been at the start of the trail.

Your face would have ended up dry, and your guide's face would have ended up with egg on it, because the graphs representing your trips up and down never cross!

Asymptotes in the middle of an interval also invalidate the IVT. For example, the function f(x) = 1/x is continuous at every point except x=0 on the interval [-1,1]. We have that f(-1) = -1 and f(1) = 1. Yet -1 < 0 < 1 and there is no c in [-1,1] for which 1/c = 0.

Equipment Check 1: Here are four statements which a shady character has justified using the Intermediate Value Theorem. Your task is to decide whether each statement is

• False
• True, but not a consequence of the IVT, or
• A valid consequence of the IVT.
As usual, check your answers with the "check answer" buttons and look at more detailed explanations using the "explain" buttons.

 Statement 1: If g(x) is a function with g(0) = 4 and g(3) = 7, then there is some x between 0 and 3 such that g(x) = 5. False. True, but not due to the IVT. True because of the IVT Detailed Explanation. Statement 2:Let f(x) = x3 + 2 x2 - 5. There is some point between 1 and 2 where f(x) = 4. False. True, but not due to the IVT. True because of the IVT Detailed Explanation. Statement 3: If f(t) is continuous on the closed interval [-5,3], f(-5)=6, and f(3)=-9, then f(0) could be equal to 10. False. True, but not due to the IVT. True because of the IVT Detailed Explanation. Statement 4: A function f(t) depicts the quantity of flimflam present at time t. At t=0 there were 10 cubic hands of flimflam and a week later there were 40 cubic hands of flimflam present. At no time were there exactly 20 cubic hands of flimflam present. Therefore the function f(t) is not continuous. False. True, but not due to the IVT. True because of the IVT Detailed Explanation.

Equipment Check 2: This is a "pencil and paper" equipment check. Explain why each of the following two statements does not contradict the Intermediate Value Theorem. Write out your explanations and then check an "official" explanation by clicking on the "explain" button.

1. The function sin(x) never takes on the value 2 on the closed interval [0, 2 ]. Why does this not contradict the IVT?

2. The function f(x) = (x+1)/(x-1) is zero at x = -1 and has the value 3 at x = 2. Yet there is no x for which f(x) = 1. Why does this not contradict the IVT?

View explanation.

/Stage4/Lesson/insideIVT.html