The problem you're tackling is an application of the Intermediate Value Theorem (IVT) in the context of a differentiable function, with some constraints on the derivatives at the endpoints. Here's the formal breakdown of the proof for the given statement:

Given:

• $f$ is continuous on $[a, b]$, differentiable on $(a, b)$,
• The one-sided derivatives exist: $f'(a+)$ and $f'(b-)$,
• It is assumed that $f'(a+) < f'(b-)$,
• We need to prove that for every $h \in (f'(a+), f'(b-))$, there exists some $x \in (a, b)$ such that $f'(x) = h$.

Step 1: Set up the function to apply the Mean Value Theorem

We aim to prove that for every $h \in (f'(a+), f'(b-))$, there exists an $x \in (a, b)$ such that $f'(x) = h$. To do this, we will adapt the proof of the Mean Value Theorem (MVT).

Let’s define a new function $g$ based on $f$:

$g(t) = f(t) - ht$

where $h$ is some number between $f'(a+)$ and $f'(b-)$. Note that $g(t)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and $h$ is a constant.

Step 2: Apply the Mean Value Theorem

Now, apply the Mean Value Theorem (MVT) to the function $g(t)$ on the interval $[a, b]$. The MVT tells us that there exists some $c \in (a, b)$ such that:

$g'(c) = \frac{g(b) - g(a)}{b - a}$

Let’s compute $g'(t)$:

$g'(t) = f'(t) - h$

Using MVT on $g$, we obtain:

$f'(c) - h = \frac{g(b) - g(a)}{b - a}$

Note that the expression on the right-hand side is a constant. So, for some $c \in (a, b)$, we have:

$f'(c) - h = 0 \quad \Rightarrow \quad f'(c) = h$

Thus, for every $h \in (f'(a+), f'(b-))$, there exists $c \in (a, b)$ such that $f'(c) = h$.

Step 3: Conclusion

By the construction of $g(t)$, we have shown that for every $h \in (f'(a+), f'(b-))$, there exists an $x \in (a, b)$ such that $f'(x) = h$. This completes the proof.

Remark:

As a result, we have proven that the range of $f'(t)$ is an interval, possibly unbounded, even though $f'(t)$ is not necessarily continuous. This conclusion is important because the Mean Value Theorem or Intermediate Value Theorem applies even in cases where the derivative is not continuous.

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Would you like further clarification on any part of this proof? Or do you need more examples related to this concept?

Here are some related questions for you to explore:

1. How does the Intermediate Value Theorem (IVT) apply to continuous functions on closed intervals?
2. What is the connection between the Mean Value Theorem and the Fundamental Theorem of Calculus?
3. How do you prove the existence of a root for a continuous function using IVT?
4. What happens if the function $f$ is not differentiable at the endpoints?
5. How does this result extend to functions that are differentiable on an open interval but not necessarily at the endpoints?

Tip: When using the Mean Value Theorem (MVT), ensure you check the conditions carefully—particularly continuity on the closed interval and differentiability on the open interval!