Question

Can anyone explain intermediate value theorem?!

Answer 1

if you have a continuous function that have values -1 and 1 then at some point the function must evaluate to 0 as well (or any other mid-value)

Answer 2

So how do I prove that the function f(x)=x+3 has a a root in the interval [1,2]?

Answer 3

hmm, did the function and interval print correctly?

Answer 4

f(x)=4x^3-6x^2+3x-2 has a root in (1,2)

Answer 5

Well, in this case a = 1 and b = 2. I'd first plug in 1 and 2 into f(x) to find your f(a) and f(b).

Answer 6

What do you get?

Answer 7

f(a)=-1 and f(b)=12

Answer 8

Okay, sorry for late response, but since -1 < 0 < 12, we know there exists a c where 1 <c < 2 where f(c) = 0. This is basically what intermediate value theorem tells us.

Answer 9

So basically you can take anything in that interval as long as we can prove that it exists?

Answer 10

If we can prove the function is continuous on that interval (in the case of a polynomial, this is trivial), then we can use the intermediate value theorem to show

Answer 11

That if f(a) < N < f(b), then then there is a c where a<c<b and f(c) = N

Might be off on edge cases here.

Answer 12

sorry, stepped out for food. Having the function makes a lot more sense now. Yes, since the function is continuous there exists all the values between -1 and 12 someplace on the curve between x [1,2]