f(x)=2x^2-3x-9

Question

g(x)=4x^2-9
find (f/g)(x)
a)x-3/2x+3
b)1/2+1/3x+1
c)x+3/2x-3
d)x-3/2x-3

Answer 1

To find \((f/g)(x)\), we need to compute the quotient of the two functions \(f(x)\) and \(g(x)\):

1. Define the functions: \[ f(x) = 2x^2 - 3x - 9 \] \[ g(x) = 4x^2 - 9 \]

2. Calculate \((f/g)(x)\): \[ (f/g)(x) = \frac{f(x)}{g(x)} = \frac{2x^2 - 3x - 9}{4x^2 - 9} \]

3. Factor the numerator and the denominator:

   • Factor \(g(x)\): \[ g(x) = 4x^2 - 9 = (2x - 3)(2x + 3) \quad \text{(difference of squares)} \]

   • Factor \(f(x)\): We will factor \(2x^2 - 3x - 9\). To factor it, we need to find two numbers that multiply to \(2 \cdot (-9) = -18\) and add to \(-3\). These numbers are \(-6\) and \(3\). \[ f(x) = 2x^2 - 6x + 3x - 9 = 2x(x - 3) + 3(x - 3) = (2x + 3)(x - 3) \]

4. Now we can rewrite \((f/g)(x)\): \[ (f/g)(x) = \frac{(2x + 3)(x - 3)}{(2x - 3)(2x + 3)} \] The \((2x + 3)\) in the numerator and denominator cancels out, resulting in: \[ (f/g)(x) = \frac{x - 3}{2x - 3} \quad \text{(valid as long as \(x \neq -\frac{3}{2}\))} \]

5. Final Result: The simplified form is: \[ (f/g)(x) = \frac{x - 3}{2x - 3} \]

Thus, the correct answer is: d) \(\frac{x - 3}{2x - 3}\).