Show that the function $f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $c<0$ and $ c\geqslant 1$.

Question

Show that the function $f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$ attains any real value if $c<0$ and $c\geqslant 1$.

Here is what I've tried so far

Function is not defined when denominator is equal to 0.

Therefore $x^2+4x+3c \neq 0,$ and so $(x+2)^2\neq 4-3c.$

How should I proceed from here?

Answer 1

Outline . . .

The following claims can be verified algebraically:

• If $\,c > 1$, then $0\,$ is not in the range.$\\[6pt]$
• If $\,c < 0$, then $1/2\,$ is not in the range.$\\[6pt]$
• If $\,c = 0\,$ or $c=1$, the numerator and denominator have a common factor, and after simplifying, it will be evident that $1$ is not in the range.$\\[6pt]$
• If $\,0 < c < 1$, then $f$ has two vertical asymptotes, has exactly one zero between the asympotes, and changes sign from negative to positive when crossing through the $x$-axis. Hence, between the two asymptotes, $f$ approaches minus infinity near the left one, and plus infinity near the right one. It follows that $f$ has full range between the asymptotes.

Therefore $f$ has full range if and only if $\,0 < c < 1$.

Answer 2

$$f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$$

$$\begin{align} & \implies (x^2+4x+3c)f=x^2+2x+c \\[5px] & \implies(f-1)x^2+(4f-2)x+c(3f-1)=0 \tag{1} \end{align} $$

[$\cdots$]  Not sure where dxiv intended to go from here (assuming its all correct so far), so I posted as community wiki if he wants to add to it.

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[ @dxiv ]   For $(1)$ to have real roots, the discriminant of the quadratic in $x$ must be non-negative:

$$ 0 \le \frac{1}{4}\Delta_x = (2f-1)^2-(f-1)(3f-1)c = \cdots = (4-3c)f^2 - 4(1-c)f + 1 - c \tag{2} $$

For inequality $(2)$ to hold for $\forall f \in \mathbb{R}\,$, the leading coefficient must be positive and the discriminant of the quadratic in $f$ be non-positive:

$$ \begin{cases} 0 \lt 4 - 3c \\ 0 \ge \frac{1}{4} \Delta_f = 4(1-c)^2 - (4-3c)(1-c) = \cdots = -c\,(1-c) \end{cases} $$