## College Algebra (6th Edition)

Published by Pearson

# Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.6 - Page 420: 61

#### Answer

Domain: {$x | x\geq\frac{1}{2}$}

#### Work Step by Step

The square root of a negative number has no solution within the domain of Real numbers. That is to say, for all functions $f(x) = \sqrt x$, $x$ cannot be less than $0$. Knowing this, we can calculate the domain of this square root function by solving for $x\gt0$. In the exercise, therefore, we can take the function $f(x) = \sqrt{2x^{2} - 5x + 2}$ and find the domain by solving for the solution set of: $$x_{domain}: (2x^{2} - 5x + 2 \geq 0)$$ By using the quadratic formula $\frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$ we can find the zeros of the function: $$2x^{2} - 5x + 2 \geq 0$$ $$x = \frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a} = \frac{-(-5) \frac{+}{} \sqrt {(-5)^{2} - 4(2)(2)}}{2(2)} = \frac{5 \frac{+}{} \sqrt {25 - 16}}{4} = \frac{5 \frac{+}{} \sqrt {9}}{4} = \frac{5 \frac{+}{} 3}{4}$$ where $x = \frac{5 + 3}{4} = 2$ and $x = \frac{5-3}{4} = \frac{1}{2}$ and the inequality can now be re-written as: $$(x - 2)(x-\frac{1}{2})\geq0$$ Finally, we come to the conclusion that, for the domain of this function, $x\geq0$ and $x\geq\frac{1}{2}$. Since both inequalities go in the same direction regarding the number line, we can say that $x$ cannot be less than $\frac{1}{2}$ satisfies both inequalities, and as such, the domain becomes {$x | x\geq\frac{1}{2}$}.

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