Answer
a) $33$
b)$\frac{-1}{4}$
c) $f(a)=3a^2-4a+1$
d) $\frac{12}{m^2}-\frac{8}{m}+1$
e) $x=0$ or $x=\frac{4}{3}$
Work Step by Step
$f(x)=3x^2-4x+1$
$a) f(4) = 3(4)^2-4(4)+1=33$
$b) f(\frac{-1}{2})=3(\frac{1}{2})^2-4(\frac{1}{2})+1=\frac{-1}{4}$
$c) f(a)=3a^2-4a+1$
$d) f(\frac{2}{m})=3(\frac{2}{m})^2-4(\frac{2}{m})+1=\frac{12}{m^2}-\frac{8}{m}+1$
e) $f(x)=1$
$3x^2-4x+1=1$
$3x^2-4x=0$
$x=0$ or $x=\frac{4}{3}$
