Question 6
Use your graphing calculator to solve the equation graphically for all real solut $x^{3}-6 x^{2}+4 x+16=0$
Solutions: $\boldsymbol{x}=$
Make sure your answers are accurate to at least two decimals Question Help: $\square$ Video
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Step 1 ：\(f'(x) = 3x^{2}-12x+4\)

Step 2 ：\(3x^{2}-12x+4=0\)

Step 3 ：\(x^{2}-4x+\frac{4}{3}=0\)

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

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 \((-\infty, 2-\sqrt{\frac{8}{3}})\), \((2-\sqrt{\frac{8}{3}}, 2+\sqrt{\frac{8}{3}})\), and \((2+\sqrt{\frac{8}{3}}, \infty)\) in the original function

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

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