To determine the domain of the function [tex]\( f(x) = 2x^2 + 5 \sqrt{x+2} \)[/tex], we need to identify all values of [tex]\( x \)[/tex] for which the function is defined. This involves considering each part of the function separately and then finding the common values of [tex]\( x \)[/tex] that satisfy all conditions.
### Step-by-Step Solution:
1. Quadratic Term:
[tex]\[
2x^2
\][/tex]
The term [tex]\( 2x^2 \)[/tex] is defined for all real numbers since the square of any real number is a non-negative real number, and multiplying by 2 will not change the domain.
2. Square Root Term:
[tex]\[
5 \sqrt{x+2}
\][/tex]
The term [tex]\( 5 \sqrt{x+2} \)[/tex] involves a square root, which is only defined for non-negative arguments (inside the square root).
3. Finding the Non-negative Condition:
We need the expression inside the square root to be non-negative:
[tex]\[
x + 2 \geq 0
\][/tex]
To solve this inequality, isolate [tex]\( x \)[/tex]:
[tex]\[
x \geq -2
\][/tex]
This tells us that [tex]\( x \)[/tex] must be greater than or equal to [tex]\(-2\)[/tex] for the square root term to be defined.
### Conclusion:
Combining our results, the quadratic term [tex]\( 2x^2 \)[/tex] is defined for all real numbers, and the square root term [tex]\( \sqrt{x+2} \)[/tex] is defined for [tex]\( x \geq -2 \)[/tex]. Therefore, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers [tex]\( x \)[/tex] such that [tex]\( x \geq -2 \)[/tex].
The completed statement is:
The domain for [tex]\( f(x) \)[/tex] is all real numbers greater than or equal to [tex]\(-2\)[/tex].
