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

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

Answer here:

Sagot :

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

1. Identify the components of the function: The function [tex]\( f(x) \)[/tex] consists of two terms: [tex]\( 2x^2 \)[/tex] and [tex]\( 5\sqrt{x-2} \)[/tex].

2. Determine the restrictions:
- The term [tex]\( 2x^2 \)[/tex] is a polynomial and is defined for all real numbers.
- The term [tex]\( 5\sqrt{x-2} \)[/tex] involves a square root. A square root is defined only when the expression inside it is non-negative.

3. Set up the inequality for the square root term: For the square root [tex]\( \sqrt{x-2} \)[/tex], the expression inside must be greater than or equal to 0:
[tex]\[ x - 2 \geq 0 \][/tex]

4. Solve the inequality:
[tex]\[ x \geq 2 \][/tex]

5. Combine the results: The restriction from the square root term tells us that [tex]\( x \)[/tex] must be greater than or equal to 2. The polynomial term does not impose any additional restrictions.

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

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