√n is defined only for n>=0
So, the domain of √(2x+1) is (2x+1)>=0, or x >= -1/2
similarly for the other.
There is always a domain.
Fnd domain
K(x)=6square root 2x+1 -5
and
g(x)= 2x^3+ square root x+6
I think its all real numbers since there isnt a domain to determine... sooo im not sure if its right
5 answers
giggity
I'm sorry, I don't understand what you mean by "giggity". Can you please provide more context or clarify your statement?
The correct answer is X>36.7
I apologize for my previous response. Let me provide the correct domains for the given functions:
For K(x) = 6√(2x+1) - 5, the domain is (2x+1) ≥ 0, since √(2x+1) is defined only for (2x+1) ≥ 0. So, solving for x, we get:
2x + 1 ≥ 0
2x ≥ -1
x ≥ -1/2
Therefore, the domain of K(x) is x ≥ -1/2.
For g(x) = 2x^3+√(x+6), the domain is (x+6) ≥ 0, since √(x+6) is defined only for (x+6) ≥ 0. So, solving for x, we get:
x + 6 ≥ 0
x ≥ -6
Therefore, the domain of g(x) is x ≥ -6.
I apologize for any confusion caused by my previous response.
For K(x) = 6√(2x+1) - 5, the domain is (2x+1) ≥ 0, since √(2x+1) is defined only for (2x+1) ≥ 0. So, solving for x, we get:
2x + 1 ≥ 0
2x ≥ -1
x ≥ -1/2
Therefore, the domain of K(x) is x ≥ -1/2.
For g(x) = 2x^3+√(x+6), the domain is (x+6) ≥ 0, since √(x+6) is defined only for (x+6) ≥ 0. So, solving for x, we get:
x + 6 ≥ 0
x ≥ -6
Therefore, the domain of g(x) is x ≥ -6.
I apologize for any confusion caused by my previous response.
2 answers