Question
Find the domain of the function.
f left parenthesis x right parenthesis equals StartRoot 3 x squared minus 19 x plus 20 EndRoot
Question content area bottom
Part 1
The domain is
enter your response here.
(Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
f left parenthesis x right parenthesis equals StartRoot 3 x squared minus 19 x plus 20 EndRoot
Question content area bottom
Part 1
The domain is
enter your response here.
(Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)
Answers
GPT-4o mini
To find the domain of the function \( f(x) = \sqrt{3x^2 - 19x + 20} \), we need to determine where the expression inside the square root is non-negative, since the square root is only defined for non-negative values.
So, we set up the inequality:
\[
3x^2 - 19x + 20 \geq 0
\]
Next, we will find the roots of the quadratic equation \( 3x^2 - 19x + 20 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = -19 \), and \( c = 20 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-19)^2 - 4 \cdot 3 \cdot 20 = 361 - 240 = 121
\]
Since the discriminant is positive, we will have two distinct real roots. Now, calculate the roots:
\[
x = \frac{19 \pm \sqrt{121}}{2 \cdot 3} = \frac{19 \pm 11}{6}
\]
Calculating the roots:
1. For \( 19 + 11 = 30 \):
\[
x = \frac{30}{6} = 5
\]
2. For \( 19 - 11 = 8 \):
\[
x = \frac{8}{6} = \frac{4}{3}
\]
Thus, the roots are \( x = 5 \) and \( x = \frac{4}{3} \).
Next, we analyze the quadratic \( 3x^2 - 19x + 20 \) to determine its sign across the intervals defined by these roots:
- The intervals are \( (-\infty, \frac{4}{3}) \), \( \left(\frac{4}{3}, 5\right) \), and \( (5, \infty) \).
- The parabola opens upwards (since the coefficient of \( x^2 \) is positive).
To find out where \( 3x^2 - 19x + 20 \) is non-negative, we can check the signs of the quadratic in each interval:
1. For \( x < \frac{4}{3} \) (e.g., \( x = 0 \)):
\[
3(0)^2 - 19(0) + 20 = 20 \quad (\text{positive})
\]
2. For \( \frac{4}{3} < x < 5 \) (e.g., \( x = 4 \)):
\[
3(4)^2 - 19(4) + 20 = 48 - 76 + 20 = -8 \quad (\text{negative})
\]
3. For \( x > 5 \) (e.g., \( x = 6 \)):
\[
3(6)^2 - 19(6) + 20 = 108 - 114 + 20 = 14 \quad (\text{positive})
\]
Thus, the function \( f(x) \) is non-negative (i.e., defined) outside the interval \( \left(\frac{4}{3}, 5\right) \) and at the points \( \frac{4}{3} \) and \( 5 \).
Therefore, the domain of the function in interval notation is:
\[
(-\infty, \frac{4}{3}] \cup [5, \infty)
\]
So, we set up the inequality:
\[
3x^2 - 19x + 20 \geq 0
\]
Next, we will find the roots of the quadratic equation \( 3x^2 - 19x + 20 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 3 \), \( b = -19 \), and \( c = 20 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-19)^2 - 4 \cdot 3 \cdot 20 = 361 - 240 = 121
\]
Since the discriminant is positive, we will have two distinct real roots. Now, calculate the roots:
\[
x = \frac{19 \pm \sqrt{121}}{2 \cdot 3} = \frac{19 \pm 11}{6}
\]
Calculating the roots:
1. For \( 19 + 11 = 30 \):
\[
x = \frac{30}{6} = 5
\]
2. For \( 19 - 11 = 8 \):
\[
x = \frac{8}{6} = \frac{4}{3}
\]
Thus, the roots are \( x = 5 \) and \( x = \frac{4}{3} \).
Next, we analyze the quadratic \( 3x^2 - 19x + 20 \) to determine its sign across the intervals defined by these roots:
- The intervals are \( (-\infty, \frac{4}{3}) \), \( \left(\frac{4}{3}, 5\right) \), and \( (5, \infty) \).
- The parabola opens upwards (since the coefficient of \( x^2 \) is positive).
To find out where \( 3x^2 - 19x + 20 \) is non-negative, we can check the signs of the quadratic in each interval:
1. For \( x < \frac{4}{3} \) (e.g., \( x = 0 \)):
\[
3(0)^2 - 19(0) + 20 = 20 \quad (\text{positive})
\]
2. For \( \frac{4}{3} < x < 5 \) (e.g., \( x = 4 \)):
\[
3(4)^2 - 19(4) + 20 = 48 - 76 + 20 = -8 \quad (\text{negative})
\]
3. For \( x > 5 \) (e.g., \( x = 6 \)):
\[
3(6)^2 - 19(6) + 20 = 108 - 114 + 20 = 14 \quad (\text{positive})
\]
Thus, the function \( f(x) \) is non-negative (i.e., defined) outside the interval \( \left(\frac{4}{3}, 5\right) \) and at the points \( \frac{4}{3} \) and \( 5 \).
Therefore, the domain of the function in interval notation is:
\[
(-\infty, \frac{4}{3}] \cup [5, \infty)
\]