Could someone help me to do this hard problem?I'd appreciate it!
Prove that if f(x) is continuous and one-to-one on the closed interval [a, b], then f(x) is stictly monotone on [a, b]. (Recall that strictly monotone on [a, b] means that either for all x and y such that x ∈ [a, b], y ∈ [a, b] and x < y, we have f(x) < f(y), or for all x and y such that x ∈ [a, b], y ∈ [a, b] and x < y, we have f(x) > f(y).)
Hint: Suppose it is not (this is probably the hardest part, negating the statement "$f$ is monotone").
Then, use the intermediate value theorem.
$f(a) \ne f(b)$ because the function is one-to-one.
Case 1: Assume $f(a) < f(b)$.
Let $a < x < y < b$. If $f(x) \le f(a) < f(b)$ then by the intermediate value theorem there is a $x: x < c < b$ so that $f(c) = f(a)$. That's impossible because $f$ is 1-1. So $f(a) < f(x)$ and....
... can you finish this?...
If not, I'll come back and type some more in a few hours after dinner.
