Question

Intermediate Value Theorem

Answer 1

need help understanding this theorem

Answer 2

What two conditions must be true to verify the applicability of the Intermediate Value Theorem?
1) the function must be continuous on the closed interval [a,b]
2) L is a number between f(a) and f(b)

Answer 3

I am working on about 3 problems. The directions are to determine if the theorem applies or not and state why or why not. If the theorem applies then I am to find the value guaranteed to exist by the theorem.

Answer 4

\[f(x)=\frac{ x-3 }{ x+2 }\]

Answer 5

interval given is [-1, 3]

Answer 6

sorry, I am also, given that f(c)=2/3

Answer 7

I solved for c and found that it is equal to 13

Answer 8

13 is not in the given interval of [-1,3]

Answer 9

and the function is not continuous at x=-2 since it has a vertical asymptote but from
[-1,3] the function is continuous so therefore the theorem applies, right?

Answer 10

Yes, the theorem applies because it is continuous over [-1, 3]

Answer 11

same function but I am given [-4,1], therefore theorem does not apply, correct?

Answer 12

Yes.

Answer 13

does the solution have to be in the given interval as well?

Answer 14

Are you referring to that 13?

Answer 15

yes

Answer 16

Well. The function is continuous over some neighborhood around 13. I'm not sure why they would give you that interval to consider when f(c) = 2/3 though...

Answer 17

Maybe my two conditions are incorrect.

Answer 18

What two conditions must be true to verify the applicability of the Intermediate Value Theorem?
1) the function must be continuous on the closed interval [a,b]
2) L is a number between f(a) and f(b)

Answer 19

Hmm. I think your conditions are correct. But, notice how f(-1) = -4 and f(3) = 0. Thus, f(c) is not an element in the interval between f(-1) and f(3)

Answer 20

so I guess the Theorem does not apply since 13 is not in the given interval.

Answer 21

2/3 is in the given range but 13 is not in the given domain

Answer 22

wait, 2/3 is not in the given range. The range is -4 to 0

Answer 23

Yes. Sorry for the confusion. By your conditions, we would say that when considering f(c) = 2/3 over the interval [-1, 3] , that condition 2 fails.

Answer 24

Yes!

Answer 25

but f(c)=2/3 was given

Answer 26

ok so (2) fails while (1) holds

Answer 27

Yes.

Answer 28

whereas in the interval [-4, 1] and the same function
condition 1 fails

Answer 29

Yes. Does condition 2 hold?

Answer 30

it fails but I only have to prove they are both true. Once condition 1 fails, then the THM does not apply at all.

Answer 31

Correct, I was just wondering out of curiosity.

Answer 32

ok just to make sure I am on track can you double check one more

Answer 33

Sure

Answer 34

function is x/(x-2) interval is [-1,1] and f(c) =-1/2

Answer 35

I am saying it IVT applies the guarantee value is 2/3 which is located in the given interval and f(-1)=1/3 and f(1)=-1 and y value is included in the given range

Answer 36

Yes. That's correct. You could say the the image of 2/3, -1/2, is a member of the interval [-1,1].

Answer 37

Thanks I haven't seen image since college days. I have to use Domain and Range terminology (High school level). Thanks I feel a little better about this theorem.

Answer 38

You're welcome. Its a powerful idea and comes up a lot. Good luck in your studies.

Answer 39

Thanks, I am just doing this for fun. Studying calculus again just to past time and improve my skills.

Answer 40

That's awesome. I'm doing the same.