Question 6
Use your graphing calculator to solve the equation graphically for all real solut $x^3-6x^2+4x+16=0$
Solutions: $\mathit{x}=$
Make sure your answers are accurate to at least two decimals Question Help: $◻$ Video
Submit Question

Step 1 ：$f^{\prime }(x)=3x^2-12x+4$

Step 2 ：$3x^2-12x+4=0$

Step 3 ：$x^2-4x+\frac{4}{3}=0$

Step 4 ：$x=\frac{4\pm \sqrt{(-4{)}^2-4\ast 1\ast \frac{4}{3}}}{2\ast 1}$

Step 5 ：$x=\frac{4\pm \sqrt{16-\frac{16}{3}}}{2}$

Step 6 ：$x=\frac{4\pm \sqrt{\frac{32}{3}}}{2}$

Step 7 ：$x=2\pm \sqrt{\frac{8}{3}}$

Step 8 ：Test the intervals $(-\mathrm{\infty },2-\sqrt{\frac{8}{3}})$, $(2-\sqrt{\frac{8}{3}},2+\sqrt{\frac{8}{3}})$, and $(2+\sqrt{\frac{8}{3}},\mathrm{\infty })$ in the original function

Step 9 ：The function is below the x-axis in the intervals $(-\mathrm{\infty },2-\sqrt{\frac{8}{3}})$ and $(2+\sqrt{\frac{8}{3}},\mathrm{\infty })$, and above the x-axis in the interval $(2-\sqrt{\frac{8}{3}},2+\sqrt{\frac{8}{3}})$

Step 10 ：$\boxed{{\displaystyle x=2+\sqrt{\frac{8}{3}}}}$ and $\boxed{{\displaystyle x=2-\sqrt{\frac{8}{3}}}}$ are the solutions of the equation