It is the values of x, for which g(x) is defined.
No. The set of x-values are the domain for only g. This will result in a set of images, which will be g(x). This set of g(x) values are the domain of f.
what is the domain of g(x) equals square root of x plus 1? √(x+1) ≥ 0 x+1≥0 x≥-1 Domain: [-1,∞)
x
You would have been given a function for f(x) and another function for g(x). When you want to find f(g(x)), you put the function for g(x) wherever x occurs in f(x). Example: f(x)=3x+2 g(x)=x^2 f(g(x))=3(x^2)+2 I'm not sure what you mean by address domain and range. They depend on what functions you're given.
The domain is the x value
If f(x) is the inverse of g(x) then the domain of g(x) and the range of f(x) are the same.
The domain of f is x is R (if imaginary roots are permitted, and there is nothing in the question to suggest otherwise). The domain of g is R excluding x = 5 So the domain of f + g is R excluding x = 5 and the domain of f/g is R excluding x = 0
It is necessary to know the domain of x and also what the function G(y) is before it is possible to answer the question.
No. The set of x-values are the domain for only g. This will result in a set of images, which will be g(x). This set of g(x) values are the domain of f.
what is the domain of g(x) equals square root of x plus 1? √(x+1) ≥ 0 x+1≥0 x≥-1 Domain: [-1,∞)
x
You would have been given a function for f(x) and another function for g(x). When you want to find f(g(x)), you put the function for g(x) wherever x occurs in f(x). Example: f(x)=3x+2 g(x)=x^2 f(g(x))=3(x^2)+2 I'm not sure what you mean by address domain and range. They depend on what functions you're given.
-1
anything can be put into it so... (-infinity,infinity)
Let f(x) = y y = 1 + (4/x) Now replace y with x and x with y and find equation for y x = 1 + (4/y) (x-1) = (4/y) y = 4/(x-1) This g(x), the inverse of f(x) g(x)= 4/(x-1) The domain will be all real numbers except when (x-1)=0 or x=1 So Domain = (-∞,1),(1,+∞) And Range = (-∞,0),(0,+∞) f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x So we get f(g(x)) = g(f(x)) Notice the error in copying the next part of your question It should be g'(f(x)) = 1/(f'(g(x))) g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1 f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1 1/[f'(g(x))] = 1/1 = 1 g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)
All real numbers except 2
First of all, g(x) is not a proper function since, many x values can have two values for g(x). Furthermore, the answer will depend on what the domain is, but you have not bothered to share that crucial bit of information.