Finding the inverse function of: $y=3x^2-6x+1$ I am trying to find the inverse function, $f^{-1}(x)$ of this function, $y=3x^2-6x+1$ 
Normally the teacher told us to interchange the $x$ and $y$
$x=3y^2-6y+1$ and... now transpose the formula to make $y$ the subject. But this look not possible. Is there another alternative method to find the $f^{-1}(x)?$
 A: $y=3x^2-6x+1=3(x^2-2x)+1$;
$y=3(x-1)^2 -3+1=$
$3(x-1)^2-2$;
A parabola opening upwards, vertex $(1,-2)$;
Local inverse for:
1) $x\gt 1$: 
$(1/3)(y+2)=(x-1)^2$;
$x= 1+\sqrt{(1/3)(y+2)}$;
Switch:
$y=1+\sqrt{(1/3)(x+2)}$, $x \gt -2$.
Local inverse for
2) $x \lt 1$;
$-\sqrt{(1/3)(y+2)}=x-1$;
Switch:
$y=1-\sqrt{(1/3)(x+2)}$, $x \gt -2$.
Does a global inverse exist? 
Is there an inverse function for $(-\epsilon+1, \epsilon +1)$, $\epsilon >0$ ?
A: It does not have an inverse. For a function to have an inverse it has to be bijective, that is injective ( if $f(x) = f(y)$ then $x = y$ ) and surjective ( for every $\lambda$ in the codomain there is $x$ in the domain such that $f(x)=\lambda$ ). More intuitively: Imagine the graph of the function. If you draw a horizontal line it should intersect the graph precisely once. Now back to your function.
$f:\mathbb{R}\rightarrow [-2,\infty)$,  $f(x) = 3x^2-6x+1$.
This function is not injective because $f(0) = f(2) = 1$. If you were to draw a horizontal line at height 1 it would intersect the graph of the functions twice. Therefore it cannot be inverted.
But we can split it into 2 separate function which have inverses. 
Let $f_1:(-\infty,1)\rightarrow [-2,\infty)$ , $f_1(x) = f(x)$ and 
$f_2:[1,\infty)\rightarrow [-2,\infty)$, $f_2(x) = f(x)$
Those 2 functions are in fact bijective because they are monotone ( decreasing and increasing respectively ). 
$y = 3x^2-6x+1 = 3x^2-6x+3-2 = 3(x^2-2x+1)-2 = 3(x-1)^2-2 \iff \frac{y+2}{3} = (x-1)^2 \iff \sqrt \frac{y+2}{3} = |x-1|$
For $x<1$:    $f_1^{-1}(y) = 1 - \sqrt\frac{y+2}{3}$
For $x \geq 1$:    $f_2^{-1}(y) = 1+\sqrt\frac{y+2}{3} $ 
