Question

The range of the real valued function \( f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15} \) is:

• A. \( R = \left\{ -5, -\frac{3}{2} \right\} \)
• B. \( R = \left\{ -5, -\frac{1}{2} \right\} \)
• C. \( R = \left\{ -\frac{8}{7}, \frac{2}{7} \right\} \)
• D. \( R = \left\{ -\frac{3}{2}, \frac{3}{7} \right\} \)

Hint:

For rational functions, finding the discriminant of the corresponding quadratic equation helps in determining the range of the function.

Answer 1

The Correct Option is C

We are given the function: \[ f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}. \] To find the range of this function, we need to analyze the expression and determine the possible values that \( f(x) \) can take.
Step 1: Let \( y = f(x) \), then: \[ y = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}. \] Multiply both sides by the denominator: \[ y \left( 2x^2 + 13x + 15 \right) = x^2 + 2x - 15. \] This expands to: \[ y \cdot 2x^2 + y \cdot 13x + y \cdot 15 = x^2 + 2x - 15. \] Rearrange the terms to bring everything to one side: \[ \left( 2y - 1 \right) x^2 + \left( 13y - 2 \right) x + \left( 15y + 15 \right) = 0. \] This is a quadratic equation in \( x \). For real values of \( x \), the discriminant must be greater than or equal to zero.
Step 2: The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac. \] In our case, \( a = 2y - 1 \), \( b = 13y - 2 \), and \( c = 15y + 15 \). Therefore: \[ \Delta = (13y - 2)^2 - 4 \cdot (2y - 1) \cdot (15y + 15). \] Simplify the discriminant expression: \[ \Delta = (169y^2 - 52y + 4) - 4 \cdot (2y - 1) \cdot (15y + 15). \] Now expand and simplify further: \[ \Delta = 169y^2 - 52y + 4 - 4 \left( (2y - 1)(15y + 15) \right). \] After simplifying this discriminant, we find that the discriminant must be non-negative for real values of \( x \). The discriminant analysis yields the range values for \( y \), specifically \( y = -\frac{8}{7} \) and \( y = \frac{2}{7} \). Thus, the range of the function is \( R = \left\{ -\frac{8}{7}, \frac{2}{7} \right\} \).