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

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?

• Let any real value be $y$ then solve $\frac{x^2+2x+c}{x^2+4x+3c}=y$ for $x$ and show that the quadratic has real solutions in the given cases. P.S. if c<0 and c⩾ 1 You must mean "or", not "and".
– dxiv
Commented Apr 29, 2017 at 2:32
• @dxiv I posted a partial answer. If you feel like you could finish it. I'll leave it there for a while, but if no one edits it or a better answer comes along I'll delete it. Commented Apr 29, 2017 at 2:51

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 $\,0 < c < 1$.

• How can the claims be verified? @quasi Commented Apr 29, 2017 at 4:35
• For example, if $c > 1$, the numerator is always positive, hence $f$ can't be zero. Commented Apr 29, 2017 at 4:44
• If you set $f = 1/2$, the solutions are $x = \pm\sqrt{c}$, hence if $c<0$, $f$ can't have a value of $1/2$. Commented Apr 29, 2017 at 4:45
• If $c=0$ or $c=1$, just substitute each in turn, factor, and simplify. Then using the simplified expression, set it equal to $1$ and try to solve You'll see that there will be no solutions. Commented Apr 29, 2017 at 4:49
• For $\,0 < c < 1\,$, a little more work is required. Following the outline, verify each of the sub-claims for that case. If you get stuck verfifying one of the sub-claims, feel free to ask. Commented Apr 29, 2017 at 4:53

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

\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.

[ @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:

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

• @dxiv Ah I understand what you meant now, would up vote if I could lol. Commented Apr 29, 2017 at 3:13
• @dxiv Wait now I'm confused, wasn't it correct before you edited? Why does the leading coefficient have to be positive? Also surely the discriminant of the quadratic in $f$ must be $\geq 0$? Commented Apr 29, 2017 at 3:37
• It looks like for $f(x)$ to attain any value we must have $0<c<1$ - which looks correct when graphed in GeoGebra
– WW1
Commented Apr 29, 2017 at 3:40
• @mrnovice $(2)$ requires the quadratic in $f$ to be non-negative for $\forall f \in \mathbb{R}\,$. For a quadratic not to change signs, it must have no real roots (or one double root), and for it to be non-negative it requires the leading coefficient to be positive. P.S. Had the "equal" cases reversed in the latest edit, fixed now.
– dxiv
Commented Apr 29, 2017 at 3:47
• @dxiv Ah I get it now thanks :). Also it means the answer given by the OP is incorrect. Commented Apr 29, 2017 at 3:49