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.
That all depends on the functions you are given for the problem! When you add or subtract from the original function, then we obtain the new function. If you combine the functions with different domains, then the domain of the addition/subtract obeys whatever domain other function has. For instance: f(x) = 1/x and g(x) = x f(x) + g(x) = (1/x) + x = (1 + x²)/x Then, the domain is all real values except 0 since x = 0 makes the denominator zero, hence making the whole expression undefined. If x = 0, then this makes f(0) undefined.
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 = (-ā,1),(1,+ā) And Range = (-ā,0),(0,+ā) 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.