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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 1

Lesson 16: Working with the intermediate value theorem- Intermediate value theorem
- Worked example: using the intermediate value theorem
- Using the intermediate value theorem
- Justification with the intermediate value theorem: table
- Justification with the intermediate value theorem: equation
- Justification with the intermediate value theorem
- Intermediate value theorem review

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# Justification with the intermediate value theorem: table

Example justifying use of intermediate value theorem (where function is defined with a table).

## Want to join the conversation?

- If the function oscillated between f(4) and f(6) (i.e. dipped down past zero), f(x) could reach 0. So wouldn't this still be a possible case? Or is IVT just talking about the
**necessity**of reaching a value (i.e. it's not**necessary**that f(x) = 0 in that interval, as Sal drew)(25 votes)- The question is asking whether the function
*has to*reach 0 on that interval. Even though it's possible, as you describe, it's not guaranteed by IVT, since 0 isn't between f(4) and f(6).(32 votes)

- I don't understand why the intermediate value theorem would not apply for the first question. Couldn't the function dip down to f(x)=0 between 4 and 6, even though they equal numbers higher than 0? There could still be a continuous function that could dip down, and come back up.(3 votes)
- 𝑓(𝑥) = 0
*could*have a solution between 𝑥 = 4 and 𝑥 = 6, but we can't use the IVT to say that it*definitely*has a solution there.(18 votes)

- For the first part, between 4 and 6 on x-axis, if we draw a continuous graph going downward and then upward (like a U-shape between (4,6), then will it not be possible to draw at least one x for which f(x)=0?(5 votes)
- IVT
*only*applies to the 𝑦 values [𝑓(𝑎), 𝑓(𝑏)]. As 0 falls outside of the bounds of [3, 7] we cannot use the IVT to say that 𝑓(𝑥) = 0 has a solution between [4, 6](3 votes)

- is f(c) still valid if its equal to f(a) or f(b)(4 votes)
- Yeah.It is a closed interval. This implies that it can be equal to f(a) or f(b).(5 votes)

- In the first problem, Sal could have drawn the graph in such way that one of the y values gave 0 right? the graph could have gone back down without us knowing in that interval right? Or I'm just being dumb(1 vote)
- He could have, but the thing is that he didn't need to cross the 𝑥-axis to connect the points (4, 3) and (6, 7) so we can't say for sure that there is a zero between 𝑥 = 4 and 𝑥 = 6.

There could be, but we don't know.(6 votes)

- In the AP Calculus test(s), would I be able to use acronyms like IVT?(2 votes)
- I don't know how they actually grade the AP exams, but it's better to be on the safe side and write Intermediate Value Theorem. There's no absolute guarantee that a grader will know what IVT is.(3 votes)

- Does the intermediate value theorem apply only when we have a closed interval or rather the function just has to be continuous?(2 votes)
- Both conditions have to be true. If the first one is false, then the values at the endpoints may not exist. If the second one is false, then there may be any type of discontinuity.(3 votes)

- What exactly does justification mean in this context of calculus?(2 votes)
- In any level of mathematics, including calculus, a justification is a mathematical proof that a statement
**must**be true.

Have a blessed, wonderful day!(3 votes)

- Seeing the examples, if I understand, for the intermediate value theorem to apply for an interval [a,b], a function f need to be both continuous and increasing or decreasing in that interval, that is continuously increasing or continuously decreasing. It's only then that there can exist a c such that f(c) lies between the values f(a) and f(b), is it the case ?(2 votes)
- The function does not have to be strictly increasing or strictly decreasing in the interval, or even in any subinterval. It's enough for it to be continuous.(3 votes)

- This one is also wrong, again!

Take x between [0,4]. According to the graph, the range of f(x) is [-2, 4], which is larger than f(0)=0 and f(4)=3. Therefore, there is a solution of c within [-2, 4] such that f(c) is either <0 or >3.

This comment is not a question. It is a correction!(1 vote)- "Take x between [0,4]."

Why? The questions ask about 𝑥 ∈ [4, 6] and 𝑥 ∈ [2, 4].

"According to the graph, the range of f(x) is [-2, 4], which is larger than f(0)=0 and f(4)=3."

The graph that Sal drew is only one possible graph.

All we can say about 𝑓(𝑥) over 𝑥 ∈ [0, 4] is that it takes on all values between 𝑓(2) = −2 and 𝑓(4) = 3.

𝑓(𝑥) could take on values less than −2 or greater than 3, but not necessarily.

"...There is a solution of c within [-2, 4] such that f(c) is either <0 or >3."

There is a 𝑐 ∈ [−2, 4] such that 𝑓(𝑐) < 0,

but not necessarily such that 𝑓(𝑐) > 3.

Either way, this has nothing to do with the questions asked.(2 votes)

## Video transcript

- [Instructor] The table
gives selected values of the continuous function f. All right, fair enough. Can we use the intermediate value theorem to say that the equation f
of x equal is equal to zero has a solution where four is less than or equal to x
is less than or equal to six? If so, write a justification. So, pause this video and see
if you can think about this on your own before we do it together. Okay, well let's just
visualize what's going on and visually think about the
intermediate value theorem. So, if that's my y-axis there and then let's say that this is my x-axis right over here. We've been given some points over here. We know when x is equal to
zero, f of x is equal to zero. Let me draw those. So, we have that point. When x is equal to two, y or f of x, y equals f of x is gonna be equal to a negative two. So, we have a negative
two right over there. When x is equal to four, so, three, four, f of x is equal to three. One, two, three. I'm doing it on a slightly different scale so that I can show everything. And when x is equal to six, so, five, six, f of x is equal to seven. Three, four, five, six, seven. So, right over here. Now, they also tell us that
our function is continuous. So, one intuitive way of
thinking about continuity is I can connect all of these dots without lifting my pencil. So, the function might look, I'm just gonna make up some stuff, it might look something, anything like what I just drew just now. And it could have even wilder fluctuations but that is what my f looks like. Now, the intermediate value theorem says hey, pick a closed interval. And here, we're picking
the closed interval from four to six, so let me look at that. So, this is one, two, three, four here, this is six here, so we're gonna look at
this closed interval. And the intermediate
value theorem tells us that look, if we're continuous
over that closed interval, our function f is gonna
take on every value between f of four, which in this case, so, this is f of four, is equal to three, and f of six, which is equal to seven. f of six, which is equal to seven. And so, if someone said hey,
is there gonna be a solution to f of x is equal to, say,
five over this interval? Yes. Over this interval, for some x, you're going to have f
of x being equal to five. But they're not asking us for an f of x equaling something
between these two values. They're asking us for
an f of x equaling zero. Zero isn't between f of four and f of six, and so we cannot use the
intermediate value theorem here. And so, if we wanted to write it out, we could say f is continuous but zero is not between f of four and f of six. So, the intermediate value
theorem does not apply. All right, let's do the second one. So here they say, can we use
the intermediate value theorem to say that there is a value
c such that f of c equals zero and two is less than or equal to c is less than or equal to four? If so, write a justification. We are given that f is continuous,
so let write that down. We are given that f is continuous, and if you wanna be over that interval, but they're telling us
it's continuous in general. And then we can just
look at what is the value of the function at these end points? Our interval goes from two to four, so we're talking about this
closed interval right over here. We know that f of two is going to be equal to negative two. We see it on that table. And what's f of four? f of four is equal to three. So, zero is between f of two and f of four. And you can see it visually here. There's no way to draw between
this point and that point without picking up your pen,
without crossing the x-axis, without having to point where your function is equal to zero. And so, we can say according to the
intermediate value theorem, there is a value c such that f of c is equal to zero and two is less than or equal to c is less than or equal to four. So, all we're saying is hey,
there must be a value c, and the way I drew it here,
that c value is right over where c is between two and four, where f of c is equal to zero. And this seems all mathy and a little bit confusing sometimes but it's saying something very intuitive. If I had to go from
this point to that point without picking up my pen,
I am going to at least cross every value between f of two
and f of four at least once.