Given f(X)=the square root of x-2 and g(x_=6/x-3 (+1), write the composite function g(f(x)) and state its domain. Where is the resulting function defined?
To find the composite function g(f(x)), we need to substitute f(x) into g(x).
Starting with f(x) = √(x - 2), we will substitute this into g(x).
The given function g(x) = 6/(x - 3) + 1.
Replacing x with f(x), we get:
g(f(x)) = 6/(f(x) - 3) + 1
Now, let's substitute f(x) = √(x - 2) back into g(f(x)):
g(f(x)) = 6/(√(x - 2) - 3) + 1
This is the composite function g(f(x)). Let's determine its domain.
The domain of g(f(x)) is the set of all x-values for which the expression inside the square root (√) is positive and the expression inside the denominator is not zero. Let's analyze both conditions:
1. Expression inside the square root (√):
For √(x - 2) to be real, (x - 2) must be greater than or equal to zero.
So, x - 2 ≥ 0 (x - 2 is non-negative)
Solving this inequality, we get x ≥ 2.
2. Expression inside the denominator:
We need to make sure the denominator, (√(x - 2) - 3), is not zero.
If √(x - 2) - 3 = 0, then √(x - 2) = 3, which implies x - 2 = 9.
Solving x - 2 = 9, we get x = 11.
So, the domain of g(f(x)) is x ≥ 2, excluding x = 11.
To find the composite function g(f(x)), we need to substitute f(x) into the function g(x).
Let's start by finding f(x):
f(x) = √(x-2)
Next, we substitute f(x) into g(x):
g(f(x)) = 6/(f(x)-3) + 1
Since f(x) = √(x-2), we substitute it into g(x):
g(f(x)) = 6/(√(x-2)-3) + 1
Now, let's simplify the expression further:
To simplify the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is (√(x-2)+3):
g(f(x)) = [6/(√(x-2)-3)] * [(√(x-2)+3)/(√(x-2)+3)]
Multiplying the numerators and denominators, we get:
g(f(x)) = [6(√(x-2)+3)] / [(√(x-2))^2 - 3^2]
Simplifying the denominator further:
g(f(x)) = [6(√(x-2)+3)] / [x - 2 - 9]
g(f(x)) = [6(√(x-2)+3)] / (x - 11)
Now, let's determine the domain of the resulting function, g(f(x)):
The resulting function is defined when the denominator is not equal to zero since division by zero is undefined. Therefore, the domain of g(f(x)) is all real numbers except x = 11.
In interval notation, the domain is (-∞, 11) U (11, ∞).