To determine the domain of the function
\[ f(x) = \frac{x^2 - 16}{3x^3 - 11x^2 - 4x}, \]
we need to find where the function is undefined, which occurs when the denominator equals zero.
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Find where the denominator is zero: \[ 3x^3 - 11x^2 - 4x = 0 \]
We can factor out an \( x \): \[ x(3x^2 - 11x - 4) = 0 \] Thus, \( x = 0 \) is one solution. Next, we need to factor or use the quadratic formula to find the roots of \( 3x^2 - 11x - 4 \).
The quadratic formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3, b = -11, c = -4 \). Calculate the discriminant: \[ b^2 - 4ac = (-11)^2 - 4 \cdot 3 \cdot (-4) = 121 + 48 = 169. \] Since \( 169 = 13^2 \) is a perfect square, we can find the roots: \[ x = \frac{11 \pm 13}{6}. \] This gives us two fractions: \[ x = \frac{24}{6} = 4 \quad \text{and} \quad x = \frac{-2}{6} = -\frac{1}{3}. \]
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Points of discontinuity: The function is undefined (denominator equals zero) at:
- \( x = 0 \)
- \( x = 4 \)
- \( x = -\frac{1}{3} \)
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Determine the domain: The domain will be all real numbers except the points where the function is undefined. Thus, the domain in interval notation is: \[ (-\infty, -\frac{1}{3}) \cup (-\frac{1}{3}, 0) \cup (0, 4) \cup (4, \infty). \]
This means we are looking for the option:
\[ (-\infty, -\frac{1}{3}) \cup (-\frac{1}{3}, 0) \cup (0, 4) \cup (4, \infty). \]
Among the given options, the correct one is:
left parenthesis negative infinity comma negative one third right parenthesis Union left parenthesis negative one third comma 0 right parenthesis Union left parenthesis 0 comma 4 right parenthesis Union left parenthesis 4 comma infinity right parenthesis
1 answer