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.

$(2t-1)^2-(t-1)(3tc-c) \geq 0\implies 4t^2+1-4t-(3t^2c-4tc+c)\geq 0\implies t^2(4-3c)+4(c-1)t+(1-c)\geq 0$
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 $0\leq c\leq 1\implies 0\leq c\leq 1$ 
