To determine the domain of the function \( f(x) = \frac{1}{5x^2 + 23x + 12} \), we need to find the values of \( x \) for which the denominator is not zero, since division by zero is undefined.
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Set the denominator equal to zero: \[ 5x^2 + 23x + 12 = 0 \]
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Use the quadratic formula: The quadratic formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 5 \), \( b = 23 \), and \( c = 12 \).
First, calculate the discriminant: \[ D = b^2 - 4ac = 23^2 - 4 \cdot 5 \cdot 12 = 529 - 240 = 289 \] Since \( D \) is positive, there are two distinct real roots.
Now, substitute into the quadratic formula: \[ x = \frac{-23 \pm \sqrt{289}}{2 \cdot 5} = \frac{-23 \pm 17}{10} \]
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Calculate the two roots:
- First root: \[ x_1 = \frac{-23 + 17}{10} = \frac{-6}{10} = -\frac{3}{5} \]
- Second root: \[ x_2 = \frac{-23 - 17}{10} = \frac{-40}{10} = -4 \]
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Interpreting the roots: The function \( f(x) \) is undefined at \( x = -\frac{3}{5} \) and \( x = -4 \). Therefore, the domain of \( f \) is all real numbers except these two points.
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Writing the domain: In interval notation, the domain of \( f \) is: \[ (-\infty, -4) \cup (-4, -\frac{3}{5}) \cup (-\frac{3}{5}, \infty) \]
Based on this analysis, the response matches the option: \[ \text{left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative 4 comma negative start fraction 3 over 5 end fraction right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis \]
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