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

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

Answer here: _______



Answer :

Let's determine the domain for the function [tex]\( f(x) = 2x^2 + 5\sqrt{x-2} \)[/tex].

1. Identify the components of the function:
- The term [tex]\( 2x^2 \)[/tex] is a polynomial, and polynomials are defined for all real numbers.
- The term [tex]\( 5\sqrt{x-2} \)[/tex] contains a square root, and the expression inside the square root must be non-negative (i.e., [tex]\( x - 2 \geq 0 \)[/tex]).

2. Set the inequality for the square root:
- To ensure the expression under the square root is non-negative, we need [tex]\( x - 2 \geq 0 \)[/tex].
- Solving this gives [tex]\( x \geq 2 \)[/tex].

3. Combine the results:
- The polynomial part [tex]\( 2x^2 \)[/tex] does not affect the domain since it's defined for all real numbers.
- The square root part mandates that [tex]\( x \geq 2 \)[/tex].

Thus, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to 2.

Therefore, the complete statement is:

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