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Justification with the intermediate value theorem: table

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.A (LO)
,
FUN‑1.A.1 (EK)
Example justifying use of intermediate value theorem (where function is defined with a table).

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  • duskpin ultimate style avatar for user Lee
    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)
    (20 votes)
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  • aqualine ultimate style avatar for user Molly S
    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.
    (2 votes)
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  • blobby green style avatar for user awoelmi321
    is f(c) still valid if its equal to f(a) or f(b)
    (3 votes)
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  • piceratops ultimate style avatar for user Matthew Chen
    In the AP Calculus test(s), would I be able to use acronyms like IVT?
    (1 vote)
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  • leaf red style avatar for user Tapiwa Hellcat
    Does the intermediate value theorem apply only when we have a closed interval or rather the function just has to be continuous?
    (1 vote)
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  • sneak peak purple style avatar for user Vector Inc.
    What exactly does justification mean in this context of calculus?
    (1 vote)
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  • blobby green style avatar for user Noor
    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 ?
    (1 vote)
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  • female robot grace style avatar for user Paloma Muñoz
    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
    (0 votes)
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  • blobby green style avatar for user Qi Zhang
    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!
    (0 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      "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.
      (1 vote)

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.