Determine the roots of the function: $g(x)=2 x^{3}-2 x+1$

Step 1 ：Given the function \(g(x)=2x^3-2x+1\), we need to find the roots of the function by solving the equation \(g(x)=0\).

Step 2 ：So, we have the equation \(2x^3-2x+1=0\).

Step 3 ：We can simplify this equation by dividing the entire equation by 2, which gives us \(x^3-x+\frac{1}{2}=0\).

Step 4 ：By inspection, we can see that \(x=1\) is a root of the equation because \(1^3 - 1 + \frac{1}{2} = 1 - 1 + \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0\).

Step 5 ：So, \(x=1\) is a root of the equation.

Step 6 ：To find the other roots, we can perform polynomial division or use the factor theorem to reduce the cubic equation to a quadratic equation.

Step 7 ：The cubic equation \(x^3-x+\frac{1}{2}=0\) can be written as \((x-1)(Ax^2+Bx+C)=0\) where A, B, and C are constants to be determined.

Step 8 ：Since we know that \(x=1\) is a root, we can substitute \(x=1\) into the original equation and solve for A, B, and C.

Step 9 ：Doing so, we get the system of equations: \(A+B+C=0\), \(A+2B+4C=0\), and \(A+3B+9C=\frac{1}{2}\).

Step 10 ：Solving this system of equations, we get \(A=1\), \(B=-2\), and \(C=1\).

Step 11 ：So, the cubic equation can be written as \((x-1)(x^2-2x+1)=0\).

Step 12 ：This can be further simplified to \((x-1)^3=0\).

Step 13 ：So, the only root of the equation is \(x=1\) with multiplicity 3.

Step 14 ：Therefore, the roots of the function \(g(x)=2x^3-2x+1\) are \(x=1\) with multiplicity 3.

Step 15 ：\(\boxed{x=1}\) is the final answer.