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If [tex]$f(x) = 2x^2 + 5\sqrt{(x-2)}$[/tex], complete the following statement:

The domain for [tex][tex]$f(x)$[/tex][/tex] is all real numbers greater than or equal to 2.


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.
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