Prove that the function f(x) = (x²+2x+a)/(x²+4x+3a) takes

Prove that the function $\displaystyle f(x) = \frac{x^{2}+2x+a}{x^{2}+4x+3a}$ takes all real values when $a\in\bigl(0,1\bigr)$.

Step 1: Reduce to a quadratic

Fix $y\in\mathbb{R}$; we must solve $f(x) = y$. Clearing the denominator: \[x^{2}+2x+a = y\bigl(x^{2}+4x+3a\bigr) \;\Longleftrightarrow\; \bigl(y-1\bigr)x^{2}+\bigl(4y-2\bigr)x+a\bigl(3y-1\bigr) = 0. \tag{1}\]

Step 2: The exceptional value $y = 1$

Then (1) degenerates to $2x+2a = 0$, i.e. $x = -a$, and the denominator there is

\[ a^{2}-4a+3a = a^{2}-a = a\bigl(a-1\bigr) \neq 0 \qquad\text{for } 0\lt a\lt 1. \]

So the value $1$ is attained, at $x = -a$.

Step 3: The discriminant for $y\neq1$

Now (1) is a genuine quadratic, solvable iff

\[ \Delta_{x} = \bigl(4y-2\bigr)^{2}-4\bigl(y-1\bigr)a\bigl(3y-1\bigr) = 16y^{2}-16y+4-4a\bigl(3y^{2}-4y+1\bigr) \ge 0, \]

\[ \Delta_{x} = \bigl(16-12a\bigr)y^{2}-\bigl(16-16a\bigr)y+\bigl(4-4a\bigr) = 4\,Q(y), \qquad Q(y) = \bigl(4-3a\bigr)y^{2}-4\bigl(1-a\bigr)y+\bigl(1-a\bigr). \tag{2} \]

Step 4: Show that $Q$ is positive for every $y$

Its leading coefficient $4-3a\gt 0$ for $a\lt \tfrac43$, in particular for $a\in\bigl(0,1\bigr)$. Its own discriminant is

\[ \Delta_{y} = 16\bigl(1-a\bigr)^{2}-4\bigl(4-3a\bigr)\bigl(1-a\bigr) = 4\bigl(1-a\bigr)\Bigl[4\bigl(1-a\bigr)-\bigl(4-3a\bigr)\Bigr] = 4\bigl(1-a\bigr)\bigl(-a\bigr) = -4a\bigl(1-a\bigr). \tag{3} \]

For $0\lt a\lt 1$ both $a\gt 0$ and $1-a\gt 0$, so $\Delta_{y}\lt 0$: the parabola $Q$ has no real zeros and, opening upward, is positive everywhere. Hence $\Delta_{x} = 4Q(y)\gt 0$ for every $y$, and (1) always has two distinct real roots.

Step 5: No root spoils the denominator

A root of (1) could fail to give a value of $f$ only if it also annihilated the denominator — and then, by (1), the numerator too. Subtracting, $\bigl(x^{2}+2x+a\bigr)-\bigl(x^{2}+4x+3a\bigr) = -2\bigl(x+a\bigr)$, so a common root would have to be $x = -a$; but the denominator there equals $a(a-1)\neq0$. No solution is lost.

Step 6: Verification

\[a = \tfrac12:\ \Delta_{y} = -4\cdot\tfrac12\cdot\tfrac12 = -1\lt 0,\ \text{and } Q(y) = \tfrac52y^{2}-2y+\tfrac12\gt 0 \;\;\checkmark\]

\[y = 5:\ x = -0.4299725 \text{ or } -4.0700275,\ \text{both giving } f(x) = 5 \;\;\checkmark\]

\[y = -3:\ x = -0.4037088 \text{ or } -3.0962912; \qquad y = 1:\ x = -a = -0.5 \;\;\checkmark\]