3/4 is 0.75 and 1/2 is 0.5 right ? so how is 3/4 between 1 and 1/2 ? am i just being dumb ?

think about it like this: f(1) = 1/1 = 1 f(2) = 1/2 = 0.5 3/4 = 0.75 must fall between 0.5 and 1! In fact, it is 1/(4/3), and 4/3 is clearly between the interval [1,2], since 4/3 is ~1.33.

Does the IVT work for continuous functions only?

The IVT only applies to intervals where the function is continuous, otherwise, the function might skip some intermediate value and the theorem would be false.

At 3:03, Sal said that g(x)=3/4 is between g(1)=1 and g(2)=2.I want to know how g(x)=3/4 is between them by calculation,is it like 4/3=1.33 when it shuffle from 3/4 to 4/3?

You probably have made a small mistake. At 3:03, Sal said that *g(2)=1/2*, _not g(2)=2_. g(x)=3/4 is between 1/2 and 1. Fair enough.

At 2:14 is there a specific reason why he included the x not being equal to 0? Perhaps my lack of understanding what defined means is part of my confusion.

The interval in question is from -1 to 1, which includes 0, and at 0 the function is undefined, so that's how you establish that it isn't continuous.

What do you mean 1/x is not continuous over that interval? You did not even substitute anything into the equation..

You can determine continuity in this context by simple inspection. x=0 => undefined.

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Course: AP®︎/College Calculus BC > Unit 1

Lesson 16: Working with the intermediate value theorem

Justification with the intermediate value theorem: equation

Name: Justification with the intermediate value theorem: equation
Uploaded: 2018-08-13T10:29:55Z
Description: Example justifying use of intermediate value theorem (where function is defined with an equation).

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Example justifying use of intermediate value theorem (where function is defined with an equation).

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f8tal
Posted 5 years ago. Direct link to f8tal's post “3/4 is 0.75 and 1/2 is 0....”
3/4 is 0.75 and 1/2 is 0.5 right ? so how is 3/4 between 1 and 1/2 ? am i just being dumb ?
Button navigates to signup pageComment on f8tal's post “3/4 is 0.75 and 1/2 is 0....”
(15 votes)
Answer
- Redmond Riddell
  Posted 4 years ago. Direct link to Redmond Riddell's post “think about it like this:...”
  think about it like this:
  f(1) = 1/1 = 1
  f(2) = 1/2 = 0.5
  3/4 = 0.75 must fall between 0.5 and 1!
  In fact, it is 1/(4/3), and 4/3 is clearly between the interval [1,2], since 4/3 is ~1.33.
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  (23 votes)
Peterson Torio
Posted 4 years ago. Direct link to Peterson Torio's post “Does the IVT work for con...”
Does the IVT work for continuous functions only?
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(4 votes)
Answer
- KLaudano
  Posted 4 years ago. Direct link to KLaudano's post “The IVT only applies to i...”
  The IVT only applies to intervals where the function is continuous, otherwise, the function might skip some intermediate value and the theorem would be false.
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  (7 votes)
Saad Syed
Posted 5 years ago. Direct link to Saad Syed's post “At 3:03, Sal said that g(...”
At
3:03
, Sal said that g(x)=3/4 is between g(1)=1 and g(2)=2.I want to know how g(x)=3/4 is between them by calculation,is it like 4/3=1.33 when it shuffle from 3/4 to 4/3?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Yan
  Posted 2 years ago. Direct link to Yan's post “You probably have made a ...”
  You probably have made a small mistake. At
  3:03
  , Sal said that g(2)=1/2, not g(2)=2.
  g(x)=3/4 is between 1/2 and 1. Fair enough.
  Button navigates to signup page
  (2 votes)
Stanislav Rushnitski
Posted 5 years ago. Direct link to Stanislav Rushnitski's post “At 2:14 is there a specif...”
At
2:14
is there a specific reason why he included the x not being equal to 0? Perhaps my lack of understanding what defined means is part of my confusion.
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(2 votes)
Answer
- André Alessi
  Posted 5 years ago. Direct link to André Alessi's post “The interval in question ...”
  The interval in question is from -1 to 1, which includes 0, and at 0 the function is undefined, so that's how you establish that it isn't continuous.
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  (4 votes)
David Anaya
Posted 5 years ago. Direct link to David Anaya's post “I think something misses ...”
I think something misses to the answer at
0:36
. What would you do if you do happen to be able to demonstrate it through the intermediate value by selecting a shorter range in which there are not discontinuities. Example: with that same equation, in the case of being asked a solution for f(c) = 3/4 between -1 <= c <= 2, you could just "cut out" the portion of the function that interests you, one without discontinuities, such as 1 <= c <= 2 and demonstrate it.
Button navigates to signup pageComment on David Anaya's post “I think something misses ...”
(3 votes)
Answer
33liaze
Posted a year ago. Direct link to 33liaze's post “What do you mean 1/x is n...”
What do you mean 1/x is not continuous over that interval? You did not even substitute anything into the equation..
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(2 votes)
Answer
- cossine
  Posted a year ago. Direct link to cossine's post “You can determine continu...”
  You can determine continuity in this context by simple inspection.
  
  x=0 => undefined.
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  (2 votes)
Scott Margolin
Posted 7 months ago. Direct link to Scott Margolin's post “This notation was really ...”
This notation was really confusing, at least for me. When I see "the equation g(x) = 3/4", I just think it denotes a straight line where y = 3/4. I didn't realize it was a potential answer to the original g(x) at the top until I read the comments. Is there a less ambiguous way of expressing this, or do I just need to get better at reading these things?
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(2 votes)
Answer
- Venkata
  Posted 7 months ago. Direct link to Venkata's post “It's all in the wording. ...”
  It's all in the wording. If g(x) = 3/4 was given without context then yes, I'd think of a line passing through y = 3/4 too. But here, the question asks for a solution to g(x) = 3/4 given g(x) = 1/x. So, it means that you're given a function's value (3/4) at a point (which is unknown) and you're asked to find if IVT can be used to determine whether there is some x for which 1/x = 3/4 with the x being between 1 and 2, and Sal goes onto prove the same.
  Comment on Venkata's post “It's all in the wording. ...”
  (1 vote)
feromario99
Posted 24 days ago. Direct link to feromario99's post “How is 3/4 between 1 and ...”
How is 3/4 between 1 and 1/2 if 3/4 is greater than 1/2 ?
Am I missing something??
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(1 vote)
Answer
- kubleeka
  Posted 23 days ago. Direct link to kubleeka's post “1/2 < 3/4 < 1 3/4 is in t...”
  1/2 < 3/4 < 1
  3/4 is in the range (1/2, 1) because it's greater than 1/2.
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  (1 vote)
HadiRECC
Posted 5 years ago. Direct link to HadiRECC's post “1:08 Wanted to fix that ...”
1:08

Wanted to fix that the function 1/x is not discontinuous at the range [-1,1], but rather only discontinuous at 0. Wrong part: [-1,1] implies discontinuity at 0 but also implies discontinuity at [-1,0)U(0,1] which is wrong.
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(1 vote)
Answer
Vivian
Posted 2 months ago. Direct link to Vivian's post “Can we use the intermedia...”
Can we use the intermediate value theorem to say the equation

f(x)= 15 has a solution where 4 <= x <= 11 ?
x 0 4 11 19
f(x) 5 12 23 37
how come f(15) has a solution where 4<= x<= 11?
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(1 vote)
Answer

Video transcript

- [Instructor] Let g of x equal one over x. Can we use the intermediate value theorem to say that there is a value c, such that g of c is equal to zero, and negative one is less than or equal to c, is less than or equal to one? If so, write a justification. So in order to even use the intermediate value theorem, you have to be continuous over the interval that you care about. And this interval that we care about is from x equals negative one to one. And, one over x is not continuous over that interval, it is not defined when x is equal to zero. And so, we could say, no, because, g of x not defined, or I could say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put in parentheses not defined, at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3/4 has a solution where 1 is less than or equal to x, is less than or equal to two? If so, write a justification. All right so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval, so we could say g of x is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined for all real numbers, such that x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero, and you could say rational functions like one over x, are continuous at all points in their domains. That's going really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, take on at the end point, or actually, these are the end points we're looking at right over here. G of one is going to be equal to one over one is one, and g of two is going to be one over two. So, 3/4 is between g of one and g of two, so by the intermediate value theorem, there must be an x that is in the interval from where it's talking about the interval from one to two, such that g of x is equal to 3/4. And so, yes, we can use the intermediate value theorem to say that the equation g of x is equal to 3/4 has a solution, and we are done.