Answer :
To determine the domain of the function [tex]\( f(x) = 2x^2 + 5\sqrt{x-2} \)[/tex], we need to consider the constraints imposed by the function.
1. Analyzing the Square Root Term:
- The square root function [tex]\( \sqrt{x-2} \)[/tex] requires that the expression inside the square root be non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
- Therefore, we set up the inequality:
[tex]\[ x - 2 \geq 0 \][/tex]
- Solving this inequality, we get:
[tex]\[ x \geq 2 \][/tex]
2. Considering the Entire Function:
- The term [tex]\( 2x^2 \)[/tex] is a polynomial and is defined for all real numbers. There are no additional constraints coming from this term.
Putting it all together, the most restrictive condition is [tex]\( x \geq 2 \)[/tex].
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to 2.
1. Analyzing the Square Root Term:
- The square root function [tex]\( \sqrt{x-2} \)[/tex] requires that the expression inside the square root be non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
- Therefore, we set up the inequality:
[tex]\[ x - 2 \geq 0 \][/tex]
- Solving this inequality, we get:
[tex]\[ x \geq 2 \][/tex]
2. Considering the Entire Function:
- The term [tex]\( 2x^2 \)[/tex] is a polynomial and is defined for all real numbers. There are no additional constraints coming from this term.
Putting it all together, the most restrictive condition is [tex]\( x \geq 2 \)[/tex].
Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to 2.
