No, don't worry about this being homework. You're genuinely confused and I'm certain that all your confusion will be immediately dispelled and replaced by wonderment and understanding if you simply had someone show you an example.
Besides, I'm the only one on this board who dislikes doing other people's homework!
Anyway, your description of the
Intermediate Value Theorem, or their properties (you called them "parts"), is kind of vague. It's hard to tell what you know about it so I will just tell you what I know and you can work that into your course (it really should all be the same).
You can find some good definitions of the
Intermediate Value Theorem right
here. I'll tell you the one I was taught (by
Michael Spivak) but you should understand that they're all equivalent (I will even leave it as an exercise for you to prove).
The
Intermediate Value Theorem states that if f is continuous on [a,b] and f(a) < 0 < f(b), then there is some x in [a,b] such that f(x) = 0.
So, lets use this theorem to prove that x + sin(x) = 1 has, at least, one solution.
Let f(t) = t + sin(t) - 1 in the interval [0,&pi/6].
f(0) = -1 < 0.
f(&pi/6) = &pi/6 + 0.5 > 0.
By the
Intermediate Value Theorem, there is an x in [0,&pi/6] such that f(x) = 0.
But f(x) = x + sin(x) - 1.
So x + sin(x) - 1 = 0.
Equivalently, x + sin(x) = 1.
QED...