<s>[INST] 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.
 [/INST] 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.
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