Respuesta :
First, we need to find the first derivative of the function
[tex]\begin{gathered} f(x)=(2x\text{ - }4)(x^2\text{ - }3) \\ f(x)=2x^3\text{ - }4x^2\text{ - }6x+12 \\ f^{\prime}(x)=6x^2\text{ - }8x\text{ - 6} \\ \\ \\ \\ \\ \\ \end{gathered}[/tex]That is the derivative, let's find which values make it 0.
[tex]\begin{gathered} 6x^{2\text{ }}\text{ - 8x - 6 = 0} \\ 6(x^2\text{ - }\frac{8}{6}x\text{ - 1\rparen = 0 } \\ \\ x^2\text{ - }\frac{8}{6}x\text{ - 1= 0 } \\ \\ \\ \\ \\ \\ \\ \end{gathered}[/tex]Then, the quadratic formula will give us the x values
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2\text{ - }4ac}}{2a} \\ \\ x=\text{ }\frac{-\lparen-8\rparen\pm\sqrt{(\text{ - 8\rparen}^2\text{ - 4\lparen6\rparen\lparen-6\rparen}}}{2(6)} \\ \\ x=\frac{8\pm\sqrt{64\text{ +144}}}{12} \\ \\ x=\frac{8\pm\sqrt{208}}{12} \\ \\ x\frac{=8\pm\sqrt{(16)(13)}}{12} \\ \\ x=\frac{8\pm4\sqrt{13}}{12} \\ \\ x=\frac{4(2\pm\sqrt{13)}}{12} \\ \\ x=\frac{2\pm\sqrt{13}}{3} \\ \\ x1=\frac{2+\sqrt{13}}{3}=\text{ 1.869} \\ \\ x2=\frac{2-\sqrt{13}}{2}=\text{ - 0.535} \\ \end{gathered}[/tex]Approximately the x values are 1.869 and - 0.535
