Let \(f(x)\) be a function continuous for all \(x\in \mathbb R \) except at \(x=0\) such that:

- \(f'(x) < 0, \forall x \in(-\infty,0) \)
- \( f'(x) > 0, \forall x \in(0,\infty) \)
- \( \lim_{x \to 0^{+}} f(x) = 2 \)
- \( \lim_{x \to 0^{-}} f(x) = 3 \)
- \( f(0) = 4 \)

And that:

\[\begin{cases} \displaystyle 2\lim_{x \to 0}f(x^3 - x^2) =\displaystyle \mu \lim_{x \to 0}f(2x^4 - x^5) \\ \displaystyle \lim_{x \to 0^{+}} \dfrac{f (-x) x^2} {\left \{\frac{1-\cos x}{\lfloor f(x) \rfloor} \right \}} = \lambda \\ \displaystyle \lim_{x \to 0^{-}} \left (\left \lfloor 3f \left(\frac{x^3 - \sin^{3}(x)}{x^4}\right) \right \rfloor - f \left(\left \lfloor \frac{\sin(x^3)}{x} \right \rfloor \right) \right) = \varphi \end{cases} \]

Find \( \mu \cdot \lambda \cdot \varphi \)

**Notations:**

\( \lfloor \cdot \rfloor \) denotes the floor function.

\( \{ \cdot \} \) denotes the fractional part function.

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