Find the domain of the function g (X) =square root of 2_square root of p solve the problem
not sure what you mean. Since g(x) really ought to include an x (and no p),
the domain of √x is x≥0
I guess you meant
g(x) = √(2-√x)
with domain (2-√x) ≥ 0 or √x ≤ 2 or x ≤ 4
or
g(x) = √2 - √x
with domain x ≥ 0
If you meant something else, try using real math notation instead of ambiguous words
It's not my job to figure out what you mean.
It's your job to say it so clearly you cannot be misunderstood.
If your question means:
√ ( 2 - √ p )
then
p must be ≥ 0
because for p < 0 , √p is complex conjugate
p must be ≤ 4
because for p > 4 , 2 - √ p is negative
In this case √ ( 2 - √ p ) is also complex conjugate
So domain is:
0 ≤ p ≤ 4
Or in interval notation:
[ 0 , 4 ]
To find the domain of the function g(x) = √(2√p), we need to determine the values that x can take without causing any issues in the function. There are two things we need to consider:
1. The radicand (the expression inside the square root) cannot be negative since the square root of a negative number is not real.
2. The value of p cannot be negative or zero, as that would also result in an undefined or imaginary solution.
Let's break it down:
1. The radicand, 2√p, must be non-negative. This means that 2√p ≥ 0.
To find the values of p that satisfy this condition, we can solve the inequality:
2√p ≥ 0
Since the square root function is always non-negative, we can divide both sides by 2:
√p ≥ 0
Now, since the square root is non-negative, we can square both sides:
p ≥ 0
So, the values of p must be greater than or equal to zero.
2. From the previous step, we found that p ≥ 0. Therefore, p cannot be negative or zero.
Therefore, the domain of the function g(x) = √(2√p) is given by p > 0, or in interval notation, (0, ∞).