Question
Find the domain of the function.
f left parenthesis x right parenthesis equals StartRoot 7 x squared minus 43 x plus 6 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 7 x squared minus 43 x plus 6 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 \( f(x) = \sqrt{7x^2 - 43x + 6} \), we need to determine for which values of \( x \) the expression inside the square root is non-negative, since the square root function is defined only for non-negative values.
Set the expression inside the square root to be greater than or equal to zero:
\[
7x^2 - 43x + 6 \geq 0
\]
Next, we'll find the roots of the quadratic equation by using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 7 \), \( b = -43 \), and \( c = 6 \).
First, we calculate the discriminant:
\[
b^2 - 4ac = (-43)^2 - 4 \cdot 7 \cdot 6 = 1849 - 168 = 1681
\]
Now plug this into the quadratic formula:
\[
x = \frac{43 \pm \sqrt{1681}}{2 \cdot 7}
\]
Calculating \( \sqrt{1681} \) gives us \( 41 \):
\[
x = \frac{43 \pm 41}{14}
\]
This results in two roots:
1. \( x = \frac{84}{14} = 6 \)
2. \( x = \frac{2}{14} = \frac{1}{7} \)
The roots of the quadratic \( 7x^2 - 43x + 6 \) are \( x = \frac{1}{7} \) and \( x = 6 \). The quadratic opens upwards (since \( a = 7 > 0 \)), so \( 7x^2 - 43x + 6 \) will be non-negative outside of the interval formed by these roots.
Thus, the intervals where \( 7x^2 - 43x + 6 \geq 0 \) is:
\[
(-\infty, \frac{1}{7}] \cup [6, \infty)
\]
Therefore, the domain of the function \( f(x) \) in interval notation is:
\[
\boxed{(-\infty, \frac{1}{7}] \cup [6, \infty)}
\]
Set the expression inside the square root to be greater than or equal to zero:
\[
7x^2 - 43x + 6 \geq 0
\]
Next, we'll find the roots of the quadratic equation by using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 7 \), \( b = -43 \), and \( c = 6 \).
First, we calculate the discriminant:
\[
b^2 - 4ac = (-43)^2 - 4 \cdot 7 \cdot 6 = 1849 - 168 = 1681
\]
Now plug this into the quadratic formula:
\[
x = \frac{43 \pm \sqrt{1681}}{2 \cdot 7}
\]
Calculating \( \sqrt{1681} \) gives us \( 41 \):
\[
x = \frac{43 \pm 41}{14}
\]
This results in two roots:
1. \( x = \frac{84}{14} = 6 \)
2. \( x = \frac{2}{14} = \frac{1}{7} \)
The roots of the quadratic \( 7x^2 - 43x + 6 \) are \( x = \frac{1}{7} \) and \( x = 6 \). The quadratic opens upwards (since \( a = 7 > 0 \)), so \( 7x^2 - 43x + 6 \) will be non-negative outside of the interval formed by these roots.
Thus, the intervals where \( 7x^2 - 43x + 6 \geq 0 \) is:
\[
(-\infty, \frac{1}{7}] \cup [6, \infty)
\]
Therefore, the domain of the function \( f(x) \) in interval notation is:
\[
\boxed{(-\infty, \frac{1}{7}] \cup [6, \infty)}
\]