Find the interval of $c$ such that the rational function $$f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$$ takes all real values $\forall$ $x\in D_f$
I tried in the following way:
Let $$y=\frac{x^2+2x+c}{x^2+4x+3c}$$ converting this to a quadratic in $x$ we get
$$(y-1)x^2+x(4y-2)+c(3y-1)=0$$ and since $x \in D_f$ this equation should have Discriminant Non negative
so
$$(4y-2)^2-4c(y-1)(3y-1) \ge 0$$ i.e.,
$$(4-3c)y^2+(4c-4)y+1-c \ge 0$$ since $\forall y \in \mathbb{R}$ the above quadratic is non negative we have two conditions
$$4-3c \gt 0$$ and
$$(4c-4)^2-4(4-3c)(1-c) \lt 0$$
$$4c^2-4c \lt 0$$ so
$c \in (0 \:\: 1)$
Is there any other approach
\begin{align} f(x) &= \frac{x^2+2x+c}{x^2+4x+3c} \\ &= \frac{x^2+4x+3c - 2x - 2c}{x^2+4x+3c} \\ &= 1-\frac{2x + 2c}{x^2+4x+3c} \end{align}
therefore $x^2+4x+3c$ cannot be zero.
$x^2+4x+3c = 0$ has no solution from discrimant
\begin{align} b^2 - 4ac &< 0 \\ 16 & < 12c \\ 3c & > 4 \\ c & > \frac{4}{3} \end{align}
however if $2x + 2c$ is also zero, it would take the value, so if $(x + c)$ is a factor of denominator $x^2 + 4x + 3c = (x + c)(x + 3)$ implies that $cx + 3x = 4x$ so $c = 1$
However at $c = 1$, $x = -3$ cannot be taken.
So $c > \dfrac{4}{3}$.