Question
Determine the domain of the function f left parenthesis x right parenthesis equals start fraction 1 over 5 x squared plus 23 x plus 12 end fraction.
(1 point)
Responses
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
Image with alt 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
left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
left parenthesis negative infinity comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma infinity right parenthesis
left parenthesis negative infinity comma negative 4 right parenthesis union left bracket negative start fraction 3 over 5 end fraction comma infinity right parenthesis
(1 point)
Responses
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
Image with alt 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
left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
left parenthesis negative infinity comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma infinity right parenthesis
left parenthesis negative infinity comma negative 4 right parenthesis union left bracket negative start fraction 3 over 5 end fraction comma infinity right parenthesis
Answers
GPT-4o mini
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.
1. **Set the denominator equal to zero**:
\[
5x^2 + 23x + 12 = 0
\]
2. **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:
\[
x = \frac{-23 \pm \sqrt{289}}{2 \cdot 5} = \frac{-23 \pm 17}{10}
\]
3. **Calculate the two roots**:
- First root:
\[
x_1 = \frac{-23 + 17}{10} = \frac{-6}{10} = -\frac{3}{5}
\]
- Second root:
\[
x_2 = \frac{-23 - 17}{10} = \frac{-40}{10} = -4
\]
4. **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.
5. **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:
\[
\text{left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative 4 comma negative start fraction 3 over 5 end fraction right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
\]
1. **Set the denominator equal to zero**:
\[
5x^2 + 23x + 12 = 0
\]
2. **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:
\[
x = \frac{-23 \pm \sqrt{289}}{2 \cdot 5} = \frac{-23 \pm 17}{10}
\]
3. **Calculate the two roots**:
- First root:
\[
x_1 = \frac{-23 + 17}{10} = \frac{-6}{10} = -\frac{3}{5}
\]
- Second root:
\[
x_2 = \frac{-23 - 17}{10} = \frac{-40}{10} = -4
\]
4. **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.
5. **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:
\[
\text{left parenthesis negative infinity comma negative 4 right parenthesis union left parenthesis negative 4 comma negative start fraction 3 over 5 end fraction right parenthesis union left parenthesis negative start fraction 3 over 5 comma infinity right parenthesis
\]