Suppose $ f $ is continuous on $ [1, 5] $ and the only solutions of the equation $ f(x) = 6 $ are $ x = 1 $ and $ x = 4 $. If $ f(2) = 8$, explain why $ f(3) > 6 $.
Alright, so here's an interesting question. They don't tell you what the function is that, they tell you it's continuous on the interval from 1-5 inclusive. And then this is an interesting thing. F uh The only solutions to f of X equals six Are one and 4. Let's start graphing. This is okay, so we've got six, maybe that's three. I'm just trying to see how big one should be. Okay, one and four, So we've got at X equals one, and X equals four. The value is six, and then I'm just going to put a dotted line here at no other point. Are we allowed to cross this dotted line? That's what this only means. The only time we're allowed to touch that line, Is that one and 4? Okay. Furthermore, we have F two equals eight. We have this point on the function and it's got to be continuous All the way from 1 to 5 in that hole in that whole interval. Okay. Um So now we're asking, well, what's the value at 3? Well um it's got to be continuous. So somehow either, you know, it can be as crazy as you want, but we've got to end up here and then we've got to end up here and we've got to end up Back down to six without crossing this line. So there is no way that the value at three could be below six and be continuous. If it were we would across this line and we haven't. Um Okay, and the fancy way of saying this is the intermediate value theorem. Uh Well, is it the no, this is not really the intermediate value, the intermediate value theorem is related, but basically we've got to be continuous. We can't cross this line, so um three has to be greater than six.