Question

For real values of $x$, the range of the function $f(x) = \dfrac{2x - 3}{2x^2 + 4x - 6}$ is

• A. $(-\infty, \dfrac{1}{8}] \cup [1, \infty)$
• B. $(-\infty, \dfrac{1}{8}] \cup [\dfrac{1}{2}, \infty)$
• C. $(-\infty, \dfrac{1}{4}] \cup [\dfrac{1}{2}, \infty)$
• D. $(-\infty, \dfrac{1}{4}] \cup [1, \infty)$

Hint:

To find the range of a rational function:
Set $y = f(x)$ and rearrange into a quadratic in $x$.
Apply the discriminant condition $\Delta \ge 0$.
Solve the resulting inequality in $y$.
Check that boundary values are actually achieved by plugging back. This method works reliably for all rational functions with $x$ appearing up to degree 2 in the denominator.

Answer 1

The Correct Option is A

The given function is \( f(x) = \dfrac{2x - 3}{2x^2 + 4x - 6} \). We aim to find the range of this function.

1. The function is a rational function where the numerator is \( 2x - 3 \) and the denominator is \( 2x^2 + 4x - 6 \).
2. To find the range, it is helpful to consider where the function is defined. The function will be undefined where the denominator is zero. Set the denominator to zero and solve for \( x \): \(2x^2 + 4x - 6 = 0\). 
3. The quadratic equation \( 2x^2 + 4x - 6 = 0 \) can be solved using the quadratic formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \( a = 2 \), \( b = 4 \), and \( c = -6 \).
4. Calculate the discriminant: \(b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64\).
5. The solutions for \( x \) are: \(x = \dfrac{-4 \pm \sqrt{64}}{4} = \dfrac{-4 \pm 8}{4}\).
6. This gives \( x = 1 \) and \( x = -3 \).
7. The function is undefined at \( x = 1 \) and \( x = -3 \), hence these points are excluded from the domain.
8. To find the range, we set \( y = \dfrac{2x - 3}{2x^2 + 4x - 6} \) and rewrite it as: \(y(2x^2 + 4x - 6) = 2x - 3\).
9. Simplifying this gives a quadratic in \( x \): \(2yx^2 + (4y - 2)x + (-6y + 3) = 0\).
10. For real \( x \), the discriminant of this equation must be greater than or equal to zero: \[ (4y - 2)^2 - 4 \cdot 2y \cdot (-6y + 3) \geq 0 \]
11. Calculate the discriminant: \[ 16y^2 - 16y + 4 - 48y^2 + 24y \geq 0 \] \[ -32y^2 + 8y + 4 \geq 0 \] \[ 8y^2 - 2y - 1 \leq 0 \]
12. The quadratic \( 8y^2 - 2y - 1 \) has roots which can be found using the quadratic formula. Solving it gives: \[ y = \dfrac{2 \pm \sqrt{4 + 32}}{16} = \dfrac{2 \pm 6}{16} \] \[ y = \dfrac{8}{16} = \dfrac{1}{2} \quad \text{and} \quad y = \dfrac{-4}{16} = -\dfrac{1}{4} \]
13. The inequality \( 8y^2 - 2y - 1 \leq 0 \) holds between the roots, thus the range of \( y \) between \( -\dfrac{1}{4} \) and \( \dfrac{1}{2} \).
14. Additionally, check behavior as \( x \to \pm \infty \): \[ \lim_{x \to \pm \infty} f(x) = 0 \] yet by exclusions from domain includes ranges outwards for limits in numerator effect.
15. Thus, the range of \( f(x) \) is: \((-\infty, \dfrac{1}{8}] \cup [1, \infty)\), corresponding to the option provided.

Therefore, the correct answer is: \((- \infty, \dfrac{1}{8}] \cup [1, \infty)\).

Answer 2

We are given: \[ f(x) = \frac{2x - 3}{2x^2 + 4x - 6}. \] Let the range be all real \(y\) such that the equation \[ y = \frac{2x - 3}{2x^2 + 4x - 6} \] has real solutions for \(x\).
Step 1: Rearranging. \[ y(2x^2 + 4x - 6) = 2x - 3. \] Expand: \[ 2yx^2 + 4yx - 6y = 2x - 3. \] Bring all terms to one side: \[ 2yx^2 + (4y - 2)x + (-6y + 3) = 0. \] For real \(x\), the discriminant must be \(\ge 0\): \[ (4y - 2)^2 - 4(2y)(-6y + 3) \ge 0. \]
Step 2: Simplify the discriminant. Compute: \[ (4y - 2)^2 = 16y^2 - 16y + 4. \] Next: \[ -4(2y)(-6y + 3) = -8y(-6y + 3) = 48y^2 - 24y. \] Thus: \[ \Delta = (16y^2 - 16y + 4) + (48y^2 - 24y) = 64y^2 - 40y + 4. \] We require: \[ 64y^2 - 40y + 4 \ge 0. \]
Step 3: Solve the quadratic inequality. Divide by 4: \[ 16y^2 - 10y + 1 \ge 0. \] The discriminant is: \[ (-10)^2 - 4 \cdot 16 \cdot 1 = 100 - 64 = 36. \] Roots: \[ y = \frac{10 \pm 6}{32}. \] Thus: \[ y_1 = \frac{16}{32} = \frac12, \qquad y_2 = \frac{4}{32} = \frac18. \] Since this quadratic opens upward (coefficient of \(y^2\) is positive), the inequality \[ 16y^2 - 10y + 1 \ge 0 \] holds for: \[ y \le \frac18 \quad \text{or} \quad y \ge \frac12. \]
Step 4: Check which endpoint values are actually attained. Evaluate whether \(y = \tfrac12\) belongs to the range. Solve: \[ \frac{2x - 3}{2x^2 + 4x - 6} = \frac12. \] Cross-multiply: \[ 2(2x - 3) = 2x^2 + 4x - 6. \] \[ 4x - 6 = 2x^2 + 4x - 6. \] Subtract LHS: \[ 0 = 2x^2. \] Thus \(x = 0\). Check denominator at \(x=0\): \[ 2(0)^2 + 4(0) - 6 = -6 \ne 0. \] Hence \(y = \frac12\) is included. Similarly for \(y = \frac18\): \[ \frac{2x - 3}{2x^2 + 4x - 6} = \frac18. \] Cross-multiply: \[ 8(2x - 3) = 2x^2 + 4x - 6. \] \[ 16x - 24 = 2x^2 + 4x - 6. \] Bring all to one side: \[ 2x^2 - 12x + 18 = 0. \] Divide by 2: \[ x^2 - 6x + 9 = 0 \Rightarrow (x - 3)^2 = 0. \] So \(x = 3\), but check the denominator at \(x = 3\): \[ 2(9) + 4(3) - 6 = 18 + 12 - 6 = 24 \ne 0, \] Hence \(y = \frac18\) is also included.
Final Range: \[ \boxed{(-\infty, \frac18] \cup [\frac12, \infty)}. \] Which matches option **(A)**.