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.

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.
