Question

Which form shows that \[f'(x)=3x^2-6x+4\] is positive for every \[x\in\mathbf{R}\]?

Options

• \[3(x-1)^2=0\]

• \[3(x-1)^2+1>0\]

• \[3(x+1)^2+1>0\]

• \[3(x-1)^2-1>0\]

Answer 1

Completing the square gives \[3x^2-6x+4=3(x-1)^2+1\]. Since \[(x-1)^2\geq0\], this expression is always greater than zero.