Find the functions a) f o g, b) g o f, c) f o f, d) g o g, and their domains.
f(x)= square root of x, g(x)= cube root of 1-x
These are my answers, but I am not sure about them and I only figured out one domain... that is the part that really confuses me. Help please!
a) 5th root of 1-x
Domain?
b) The cube root of (1 minus the square root of x)
Domain?
c) Fourth root of x
Domain= [0, infinity]
d) The cube root of (1 minus the cube root of 1 minus x)
Domain?
a. Odd roots of negative numbers can be negative, thus (-32)^1/5 has a root -2. So the domain of x is all real x.
b.same argument
c. you cannot take the even root of a negative number, so you are correct.
d. same a,b
To find the composite functions and their domains, let's evaluate each case one by one:
a) f o g: This means we substitute g(x) into f(x). So, f(g(x)) = f(cube root of (1-x)). Thus, f o g(x) = square root of (cube root of (1-x)). To find the domain, we need to identify any restrictions on x for which this function is defined. In this case, the square root function is defined for non-negative values of its argument. Therefore, the domain of f o g is x ≤ 1.
b) g o f: This means we substitute f(x) into g(x). So, g(f(x)) = cube root of (1 - square root of x). Therefore, g o f(x) = cube root of (1 - square root of x). To find the domain, we need to identify any restrictions on x for which this function is defined. In this case, both the square root and cube root functions are defined for x ≥ 0. Thus, the domain of g o f is x ≥ 0.
c) f o f: This means we substitute f(x) into f(x). So, f(f(x)) = f(square root of x) = square root of (square root of x). To find the domain, we need to identify any restrictions on x for which this function is defined. The square root function is defined for non-negative values of its argument, so the domain of f o f is x ≥ 0.
d) g o g: This means we substitute g(x) into g(x). So, g(g(x)) = g(cube root of (1 - x)) = cube root of (1 - cube root of (1 - x)). To find the domain, we need to identify any restrictions on x for which this function is defined. Both the cube root functions are defined for any real number, so there are no restrictions on the domain of g o g.
To summarize:
- a) f o g(x) = square root of (cube root of (1-x)). Domain: x ≤ 1.
- b) g o f(x) = cube root of (1 - square root of x). Domain: x ≥ 0.
- c) f o f(x) = square root of (square root of x). Domain: x ≥ 0.
- d) g o g(x) = cube root of (1 - cube root of (1 - x)). Domain: -∞ < x < ∞. (No restrictions on the domain)
Note that these domains may vary depending on any further restrictions or conditions given in the context of the problem.
Let's break it down step-by-step:
a) To find f o g, we need to substitute g(x) into f(x):
f o g(x) = f(g(x)) = f(cube root of (1-x)) = square root of (cube root of (1-x))
To determine the domain, we need to consider the restrictions of both functions. Notice that the function f(x) = square root of x is defined for x ≥ 0. Similarly, the function g(x) = cube root of (1-x) is defined for 1 - x ≥ 0 ⇒ x ≤ 1.
Therefore, the domain of f o g is x ≤ 1.
b) To find g o f, we need to substitute f(x) into g(x):
g o f(x) = g(f(x)) = g(square root of x) = cube root of (1 - (square root of x))
The domain for f(x) = square root of x is x ≥ 0. However, when we substitute f(x) into g(x), the expression (1 - square root of x) also needs to be non-negative. Therefore, the domain for g o f is x ≤ 1.
c) To find f o f, we need to substitute f(x) into f(x):
f o f(x) = f(f(x)) = f(square root of x) = square root of (square root of x)
The domain for f(x) = square root of x is x ≥ 0. When we substitute f(x) into f(x), the inner square root must also be non-negative. Hence, the domain for f o f is x ≥ 0.
d) To find g o g, we need to substitute g(x) into g(x):
g o g(x) = g(g(x)) = g(cube root of (1 - x)) = cube root of (1 - (cube root of (1 - x)))
The domain for g is x ≤ 1. Substituting g(x) into g(x) gives us the expression (1 - (cube root of (1 - x))). In this case, the inner cube root must also be defined. Hence, the domain for g o g is x ≤ 1.
Summarizing the domains:
a) f o g: x ≤ 1
b) g o f: x ≤ 1
c) f o f: x ≥ 0
d) g o g: x ≤ 1