To determine which answer is the most accurate in terms of the domain for the given function [tex]\( f(x) = \frac{\sqrt{x}}{2x^2 + x - 1} \)[/tex], we need to analyze the conditions under which the function is defined.
Let's break it down step-by-step:
1. Square Root Condition:
- The square root function [tex]\( \sqrt{x} \)[/tex] is defined only when [tex]\( x \geq 0 \)[/tex]. This means that [tex]\( x \)[/tex] must be non-negative.
2. Denominator Condition:
- The denominator [tex]\( 2x^2 + x - 1 \)[/tex] must not be zero since division by zero is undefined. To find the values of [tex]\( x \)[/tex] that make the denominator zero, solve the quadratic equation [tex]\( 2x^2 + x - 1 = 0 \)[/tex].
3. Solving the Quadratic Equation:
- Use the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2, b = 1, \)[/tex] and [tex]\( c = -1 \)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}
\][/tex]
- This yields two solutions:
[tex]\[
x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}
\][/tex]
[tex]\[
x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1
\][/tex]
- Therefore, the denominator [tex]\( 2x^2 + x - 1 = 0 \)[/tex] at [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( x = -1 \)[/tex].
4. Domain Summary:
- [tex]\( x \)[/tex] must be greater than or equal to zero.
- [tex]\( x \neq \frac{1}{2} \)[/tex] since the denominator becomes zero and the function becomes undefined at [tex]\( x = \frac{1}{2} \)[/tex].
- [tex]\( x = -1 \)[/tex] is not part of the allowable domain since it is less than zero, which does not meet the square root condition of [tex]\( x \geq 0 \)[/tex].
Given these conditions, the most accurate description of the domain is:
- [tex]\( x \geq 0 \)[/tex]
- [tex]\( x \neq \frac{1}{2} \)[/tex]
Thus, the correct answer is:
A) All Real Numbers where [tex]\( x \geq 0 \)[/tex] except [tex]\( x \)[/tex] can't equal [tex]\( \frac{1}{2} \)[/tex]
