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Justification with the intermediate value theorem: equation

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 an equation).

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  • blobby green style avatar for user f8tal
    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 ?
    (9 votes)
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  • blobby green style avatar for user Saad Syed
    At , 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?
    (3 votes)
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  • blobby green style avatar for user David Anaya
    I think something misses to the answer at . 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.
    (2 votes)
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  • old spice man green style avatar for user Stanislav Rushnitski
    At 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.
    (1 vote)
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  • leaf grey style avatar for user Peterson Torio
    Does the IVT work for continuous functions only?
    (1 vote)
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  • blobby green style avatar for user 33liaze
    What do you mean 1/x is not continuous over that interval? You did not even substitute anything into the equation..
    (1 vote)
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  • orange juice squid orange style avatar for user HadiRECC


    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.
    (0 votes)
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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.