Function problem Show that function $f(x) =\frac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$ Problem : 
Show that function $f(x)=\dfrac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $0 < c \leq 1$
My approach : 
Let the given function $f(x) =\dfrac{x^2+2x+c}{x^2+4x+3c} = t $   where $t$ is any arbitrary constant.
$\Rightarrow (t-1)x^2+2(2t-1)x+c(3t-1)=0$
The argument $x$ must be real, therefore $(2t-1)^2-(t-1)(3tc-c) \geq 0$.
Now how to proceed further? Please guide. Thanks.
 A: 1) The numerator and the denominator cannot be both zero;
2) The denominator has a zero for some $\xi\in\mathbb{R}$;
3) $\lim_{x\to\xi^\pm}f(x)=\pm\infty$;
4) $\lim_{x\to\pm\infty} f(x)=1$;
5) By continuity, $f(x)$ attains all values in $\mathbb{R}\setminus\{1\}$;
6) in $x=-\frac{1}{2}$, $f(x)=1$.
A: Note that from your quadratic equation in $x$:
$$
(t-1)x^2 + (4t-2)x^2 + (3tc-c) = 0
$$
its discriminant is:
$$
\Delta_x = (4t-2)^2-4(t-1)(3tc-c)=(-12c+16)t^2+(16 c-16)t+(-4c+4)
$$
Note that it suffices to show that $\Delta_x$ is always nonnegative. To do this, we consider its discriminant with respect to $t$:
$$
\Delta_t = (16c-16)^2-4(-12c+16)(-4c+4) = 64c(c-1)
$$
Now consider what happens when $0<c\le 1$. Observe that $\Delta_t\le0$, so the parabola $(-12c+16)t^2+(16 c-16)t+(-4c+4)$ either lies completely on one side of the $t$-axis or just barely touches the $t$-axis. But since $c\le1 \implies -12c+16\ge4$, we know that the parabola must be opening upwards so that $\Delta_x \ge 0$, as desired.
A: $(2t-1)^2-(t-1)(3tc-c) \geq 0\implies 4t^2+1-4t-(3t^2c-4tc+c)\geq 0\implies t^2(4-3c)+4(c-1)t+(1-c)\geq 0$
Now a quadratic polynomial $\geq 0$ $\forall t\in \Bbb R$  iff coefficient of second power of variable is positive and Discriminant $\leq 0$
which gives $4-3c>0\implies c<\frac{4}{3}$ and $D=16(c-1)^2+4(4-3c)(c-1)\leq 0\implies 4(c-1)(4c-4+4-3c)\leq 0\implies 4(c-1)c\leq 0\implies 0\leq c\leq 1$
So $c<\frac{4}{3}$ and $0\leq c\leq 1\implies 0\leq c\leq 1$ 
A: To continue your approach.
Rewrite $(2t−1)^2−(t−1)(3tc−c)$ as 
$(4+3c)t^2+(-4+c+3c)t+(1-c)$
This is a quadratic polynomial in $t$ with discriminant 
$(-4+4c)^2-4(1-c)(4+3c)$. You want to know that this is negative for all $c$ in the given interval. You probably know how to do this. Find the zeros of this polynomial and check the interval $(0,1)$ is inside where this polynomial is negative.
