Justification with the intermediate value theorem: table | AP Calculus AB | Khan Academy

[@a.n.onymous7198]

Is this not wrong? What's preventing f(5) = 0, with the graph between f(4) and f(6) resembling a parabola that crosses the x-axis?

[@jasonsolano2935]

nothing, but you couldn't assure that it always be the case

[@a.n.onymous7198]

@@jasonsolano2935 yeah thanks, I didn't understand what the IVT was about when I made the comment. For anyone else confused, the IVT just says that you have to cross every y-value between two given points, it doesn't say anything about points outside that range. So in this case, there's always going to be some x in [4,6] that has a corresponding y-value between 3 and 7. You can draw a parabola hitting f(x)=0, but the IVT only guarantees that the parabola has to go from 3 to 7 on the way back up, so you can't use the theorem to prove it hits f(x)=0.

[@totoblox4242]

@@a.n.onymous7198yup

[@kl-je3up]

why couldnt f(4.5) = 0? u can just draw the graph differently?

[@a.n.onymous7198]

I wrote another comment going into it, but basically the question is asking if you can use the IVT to prove it, and the IVT here only guarantees f(x) values from 3 to 7.

[@cornationboot8690]

if u want u could draw

[@mohanraina2036]

Why is that I'm the only one commenting

[@monishkm07]

Coz You have a good heart

[@tanyatripathi8929]

Because maybe people are less grateful now.

[@cornationboot8690]

cos u a wimp

[@Chill_Gates]

Because most people watched this video on the Khan Academy website and commented there rather than on KZbin