Question

I need help question

Answer 1

First, we need to find the first derivative of the function

`\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}`

That is the derivative, let's find which values make it 0.

`\begin{gathered} 6x^{2\text{ }}\text{ - 8x - 6 = 0} \\ 6(x^2\text{ - }(8)/(6)x\text{ - 1\rparen = 0 } \\ \\ x^2\text{ - }(8)/(6)x\text{ - 1= 0 } \\ \\ \\ \\ \\ \\ \\ \end{gathered}`

Then, the quadratic formula will give us the x values

`\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=(8\pm√(208))/(12) \\ \\ x(=8\pm√((16)(13)))/(12) \\ \\ x=(8\pm4√(13))/(12) \\ \\ x=(4(2\pm√(13)))/(12) \\ \\ x=(2\pm√(13))/(3) \\ \\ x1=(2+√(13))/(3)=\text{ 1.869} \\ \\ x2=(2-√(13))/(2)=\text{ - 0.535} \\ \end{gathered}`

Approximately the x values are 1.869 and - 0.535