If \(\frac{3x^2-4x+1}{x-1}=m\), and x can be any real number except 1, what real values can m NOT have?
\(f(x) = \dfrac{3x^2-4x+1}{x-1}=\\ \dfrac{(3x-1 )(x-1 )}{x-1}=3x-1\\ \text{but with a "hole" at $x=1$ so $3(1)-1=2$ is not in the range of $f(x)$}\)
\(\text{$x = 1$ is known as a removable singularity}\)
