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### Course: Differential Calculus>Unit 1

Lesson 16: Intermediate value theorem

# Justification with the intermediate value theorem: table

Discover how to apply the Intermediate Value Theorem with a table-defined function. Learn to determine if a continuous function has a solution within a specified interval, and justify your reasoning using the theorem's principles. Explore real-life applications and enhance your understanding of calculus concepts.

## 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)
• 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).
• 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.
• 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.
• 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?
• 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]
• is f(c) still valid if its equal to f(a) or f(b)
• Yeah.It is a closed interval. This implies that it can be equal to f(a) or f(b).
• 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.
• In the AP Calculus test(s), would I be able to use acronyms like IVT?
• 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.
• Does the intermediate value theorem apply only when we have a closed interval or rather the function just has to be continuous?
• 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.
• What exactly does justification mean in this context of calculus?
• In any level of mathematics, including calculus, a justification is a mathematical proof that a statement must be true.

Have a blessed, wonderful day!
• 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 ?
• 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.
• 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.