Applying MVT and IVT (?) to show something about f'''? (edit: actually, Taylor's Theorem using Lagrange remainder)

I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times. However, I can't see how to proceed. I went down what I thought were the typical paths and just reached dead ends.

Does someone maybe have a hint? I think that with the appropriate nudge I should be able to solve this question. If not, I'll edit and ask if someone knows how to solve the whole thing.

Here's the question:

Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$. Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.

We have by Taylor's theorem $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + \frac{x^{3}}{6}f'''(\xi)\tag{1}$$ for some $\xi$ between $0$ and $x$. And thus $$f(3) = \frac{9f''(0)}{2} + \frac{9}{2}f'''(\xi_{1})\tag{2}$$ for some $\xi_{1} \in (0, 3)$ and $$f(-3) = \frac{9f''(0)}{2} - \frac{9}{2}f'''(\xi_{2})\tag{3}$$ for some $\xi_{2} \in (-3, 0)$ and on subtracting these equations we get $$9 = \frac{9}{2}\{f'''(\xi_{1}) + f'''(\xi_{2})\}$$ or $$f'''(\xi_{1}) + f'''(\xi_{2}) = 2\tag{4}$$ Now it is obvious that both $f'''(\xi_{1})$ and $f'''(\xi_{2})$ can't be less than $1$ and our result follows.

Note that we don't need the continuity of $f'''$ but only its existence in the interval $(-3, 3)$. And as you can see we don't need MVT/IVT either but rather Taylor's theorem (which BTW is a consequence of the MVT).