[tex]f(x) = x^3-mx+1\\ \\ f'(x) = 3x^2-m\\ f'(x) = 0 \Rightarrow 3x^2 = m \Rightarrow x^2 = \dfrac{m}{3} \Rightarrow x = \pm \sqrt{\dfrac{m}{3}}[/tex]
[tex]\begin{array}{ccccccc}x&\Bigg|&-\infty&-\sqrt{\dfrac{m}{3}}&\,\,\,\sqrt{\dfrac{m}{3}}&+\infty&\Bigg|\\f(x)&\Bigg|&-\infty& f\left(-\sqrt{\dfrac{m}{3}}\right)&f\left(\sqrt{\dfrac{m}{3}}\right)&+\infty&\Bigg|\\\text{Sir Rolle}&\Bigg|&-&+&\,\,\,-&+&\Bigg|\end{array}[/tex]
Conform șirului lui Rolle, o soluție reală există doar între (- +) sau (+ -).
Singura posibilitate este (- x₁ + x₂ - x₃ +).
[tex]\\\boxed{1}\,\,\,\,f\left(-\sqrt{\dfrac{m}{3}}\right)> 0 \Rightarrow \left(-\sqrt{\dfrac{m}{3}}\right)^3-m\left(-\sqrt{\dfrac{m}{3}}\right)+1 > 0\\ \\ \Rightarrow -\dfrac{m\sqrt{m}}{3\sqrt{3}}+\dfrac{m\sqrt{m}}{\sqrt{3}}+1>0 \Rightarrow -m\sqrt{m}+3m\sqrt{m}+3\sqrt{3}>0\\ \\ \Rightarrow 2m\sqrt{m}>-3\sqrt{3},\quad \forall m\geq 0[/tex]
[tex]\\\boxed{2}\,\,\,\,f\left(\sqrt{\dfrac{m}{3}}\right)< 0 \Rightarrow \left(\sqrt{\dfrac{m}{3}}\right)^3-m\left(\sqrt{\dfrac{m}{3}}\right)+1 < 0\\ \\ \Rightarrow \dfrac{m\sqrt{m}}{3\sqrt{3}}-\dfrac{m\sqrt{m}}{\sqrt{3}}+1<0 \Rightarrow m\sqrt{m}-3m\sqrt{m}+3\sqrt{3}<0\\ \\ \Rightarrow -2m\sqrt{m}<-3\sqrt{3} \Rightarrow 2m\sqrt{m}>3\sqrt{3}\Big|^2\Rightarrow 4m^3>27\\ \\ \Rightarrow m^3 > \dfrac{27}{4} \Rightarrow m > \dfrac{3}{\sqrt[3]{4}}[/tex]
[tex]\\\text{Din }\,\boxed{1} \,\wedge\,\boxed{2} \,\Rightarrow \,m\in \left(\dfrac{3}{\sqrt[3]4},+\infty\right)\Rightarrow \textbf{(a) corect}[/tex]
