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
√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.
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.
To find the domain of a function, we need to identify any restrictions or limitations on the values that x can take. Let's analyze each function separately:
1. Function K(x) = 6√(2x+1) - 5:
The only potential restriction in this case is the square root term (√). Since the square root of a number is only defined for non-negative values, we need to ensure that the expression inside the square root is non-negative.
For 2x+1 ≥ 0:
2x ≥ -1
x ≥ -1/2
Therefore, the domain of K(x) is all real numbers greater than or equal to -1/2, denoted as (-1/2, ∞).
2. Function g(x) = 2x^3 + √(x+6):
Here, we again need to consider the square root term (√). The expression inside the square root must be non-negative:
For x+6 ≥ 0:
x ≥ -6
So, the function g(x) is defined for all real numbers greater than or equal to -6, denoted as [-6, ∞).
In conclusion:
- The domain of K(x) is (-1/2, ∞).
- The domain of g(x) is [-6, ∞).