IDNLearn.com connects you with a community of experts ready to answer your questions. Ask anything and receive prompt, well-informed answers from our community of knowledgeable experts.
Sagot :
To determine the domain of the function [tex]\( f(x) = 2x^2 + 5\sqrt{x - 2} \)[/tex], we need to ensure that the expression under the square root is non-negative, as the square root of a negative number is not defined in the set of real numbers.
The expression under the square root is [tex]\( x - 2 \)[/tex]. To find when this expression is non-negative, we solve the inequality:
[tex]\[ x - 2 \geq 0 \][/tex]
[tex]\[ x \geq 2 \][/tex]
Therefore, the smallest value [tex]\( x \)[/tex] can take is 2. For any [tex]\( x \)[/tex] that is less than 2, the expression [tex]\( \sqrt{x - 2} \)[/tex] would be undefined in the real number system.
Consequently, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to 2.
Thus, the complete statement is:
The domain for [tex]\( f(x) \)[/tex] is all real numbers [tex]\(\boxed{\text{greater}}\)[/tex] than or equal to 2.
The expression under the square root is [tex]\( x - 2 \)[/tex]. To find when this expression is non-negative, we solve the inequality:
[tex]\[ x - 2 \geq 0 \][/tex]
[tex]\[ x \geq 2 \][/tex]
Therefore, the smallest value [tex]\( x \)[/tex] can take is 2. For any [tex]\( x \)[/tex] that is less than 2, the expression [tex]\( \sqrt{x - 2} \)[/tex] would be undefined in the real number system.
Consequently, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to 2.
Thus, the complete statement is:
The domain for [tex]\( f(x) \)[/tex] is all real numbers [tex]\(\boxed{\text{greater}}\)[/tex] than or equal to 2.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.