I am told by my teacher that the IVT states the following:
Let f be continuous on the closed and bounded interval [a, b], and let f(a) ≤ y ≤ f(b). Then there exists a value c on [a, b] such that f (c) = y.
However, I'm confused at why f(b) is assumed to be larger than f(a). Certainly, b is larger than a, but if the function is, say, strictly decreasing…
Basically, I'm confused about the definition of the IVT.
Best Answer
If $f(b) \leq y \leq f(a)$ then also there exist $c$ such that $f(c)=y$. There is not much difference between these two cases since changing $f$ to $-f$ and $y$ to $-y$ will yield this new result. Just for definiteness they have assumed that $f(a) \leq y \leq f(b)$.