Let S = \ t ∈ R: f(x) = |x - π| ⋅ ( e ^ |x| - 1) |x|is not...

$f\left(x\right) = \left|x - \pi\right| . \left({e}^{\left|x\right|} - 1\right) . \sin \left|x\right|$

In the neighbourhood of $x = \pi$, $f\left(x\right) = \left|x - \pi\right| . \left({e}^{x} - 1\right) . \sin x$

$R{f}^{′}\left(\pi\right) = \lim_{h \to 0} \dfrac{f\left(\pi + h\right) - f\left(\pi\right)}{h} = \lim_{h \to 0}\dfrac{\left|h\right|\sin \left(\pi + h\right)\left({e}^{\pi + h} - 1\right)}{h} = 0$

$L{f}^{′}\left(\pi\right) = \lim_{h \to 0}\dfrac{f\left(\pi\right) - f\left(\pi - h\right)}{h} = - \lim_{h \to 0}\dfrac{\left|- h\right|\sin \left(\pi - h\right)\left({e}^{\pi - h} - 1\right)}{h} = 0$

Differentiable at $x = \pi$

$R{f}^{′}\left(0\right) = \lim_{h \to 0}\dfrac{f\left(h\right) - f\left(0\right)}{h} = \lim_{h \to 0}\left|h - \pi\right| \times \dfrac{\sin \left|h\right|\left({e}^{\left|h\right|} - 1\right)}{h} = 0$

$L{f}^{′}\left(0\right) = \lim_{h \to 0} \dfrac{f\left(0\right) - f\left(- h\right)}{h} = - \lim_{h \to 0}\left|- h - \pi\right| \times \dfrac{\sin \left|- h\right|\left({e}^{\left|- h\right|} - 1\right)}{h} = 0$

Differentiable at $x = 0$

Hence, $f\left(x\right)$ is differentiable everywhere.

$⇒ S = ϕ$