Every Vertical Strip Has A Point With Positive Slope Consider a differentiable function $f$ on $[x_0, x_1]$.
Suppose $f(x_0) < f(x_1)$.
It seems intuitively obvious that every vertical strip between $f(x_0)$ and $f(x_1)$ contains a point with positive slope.
i.e. a point $x$ with $f'(x) > 0$ and $a < f(x) < b$
Let the lower bound of this vertical strip be $a$.
Let the upper bound of this vertical strip be $b$.

This seems like it should be a very common question, I believe it can be proved formally using the intermediate + mean value theorem.
Does anyone have any idea how to prove this?
 A: First, judging from your drawing, you probably mean horizontal strips not vertical strips.
That aside, this problem is an application of IVT (intermediate value theorem), MVT (mean value theorem) and certain other facts leveraging the continuity of $f$ (e.g. preimages of closed sets are closed).
As $f(x_0) < a < f(x_1)$, by IVT there is some $x_0 < s < x_1$ such that $f(s) = a$. So, the set
$$
S := \{s \in [x_0, x_1] : f(s) = a\} = f^{-1}(\{a\}) \cap [x_0, x_1]
$$ is non-empty, bounded and closed. Thus, $s^\ast := \sup(S)$ is achieved inside $S$ i.e. $f(s^\ast) = a$ and $x_0 \leq s^\ast \leq x_1$ Also, $f(x_1) \neq a$. So in fact $s^\ast < x_1$.
Then, as $a = f(s^\ast) < b < f(x_1)$, yet again by IVT, there is some $s^\ast < t < x_1$ such that $f(t) = b$. So, the set
$$
T := \{t \in [s^\ast, x_1] : f(t) = b\} = f^{-1}(\{b\}) \cap [s^\ast, x_1]
$$ is non-empty, bounded and closed. Thus, $t^\ast := \inf(T)$ is achieved inside $T$ i.e. $f(t^\ast) = b$ and $s^\ast \leq t^\ast \leq x_1$. Also, $f(s^\ast) = a \neq b$. So in fact $s^\ast < t^\ast$.
Finally by MVT, there is some $s^\ast < u < t^\ast$ such that $f'(u) = \frac{f(t^\ast) - f(s^\ast)}{t^\ast - s^\ast} = \frac{b - a}{t^\ast - s^\ast} > 0$.
You must also have $a < f(u) < b$. Otherwise for e.g. if $f(u) \leq a$, then you have the situation $f(u) \leq a < f(x_1)$ so IVT would give you an $u \leq s' < x_1$ such that $f(s') = a$. But then, $s^\ast < u \leq s'$, which is impossible because $s^\ast$ is supposed to be the largest $s \in [x_0, x_1]$ such that $f(s) = a$. A similar argument shows $b \leq f(u)$ is also impossible.
