What is the domain of f(g(x))?
________________________________________... The domain of f(g(x)) is the range of g except for those values g(x) that make f undefined. Answer: [Domain of f(g(x))={x| x is a real number, and not 0, and not 5/8} ______________________________________... If g(x) = 5/(4x): (1) As x goes from negative infinity to very near 0, g(x) stays negative and goes from very near 0 to negative infinity. (2) As x goes from positive infinity to very near 0, g(x) stays positive and goes from very near 0 to positive infinity. (3) g(0) is undefined: and, therefore, does not exist. So, x cannot be 0. ______________________________ Therefore the range of g(x) is all real numbers except 0 Also, the domain of function f cannot contain 2, since f(x) = 3/sqrt(3x-6) is undefined at 2. It follows, then, that x cannot equal 5/8: because g(x)=2 when x=5/8 Therefore: [Domain of f(g(x)] = {x| x is a real number, and x not 0, and x not 5/8} ______________________________________... THE FOLLOWING IS PROBABLY THE PREFERRED METHOD THE ANSWER IS THE SAME ______________________________________... f(x)=3/sqrt(3x-6) and g(x)=5/(4x) ,,,,f(x) = 3/sqrt(3x-6) f(g(x)) = 3/sqrt[(3g(x))-6] ..........= 3/sqrt[(3(5/(4x)))-6] ..........= 3/sqrt[(15/(4x))-6] ..........= 3/sqrt[(15-24x)/(4x)] Now this is undefined only when: (1) 15-24x=0...........that's for x=5/8 (2) 4x=0..................that's for x=0 Therefore, [the domain of f(g(x))] = {x| x is a real number, and not 0, and not 5/8} ______________________________________...
Let f (x) = 1/(x+1)?.(a) show that it is invertible and find its inverse.(b) find the domain and range of f.?
a) f(x) = 1/(x+1) =y y(x+1) =1 x+1 = 1/y x = g(y) = (1-y)/y f^-1(x) = g(x) = (1-x)/x b) Domain of f(x) = all real x except x = -1. Range of f(x) = all real numbers except f(x) = 0
ONE MORE! What is the DOMAIN OF f(g(x)) if...?
The domain of g(x) is all real numbers. The domain of f(x) is all real numbers greater than or equal to zero. You can basically ignore the domain of g(x), since its domain is all real numbers. (If you had to restrict any numbers for the domain of g(x), then you would need to take that into account). The range of g(x) comes into play here because you are taking numbers from the range of g(x) and plugging them into f(x). The range of g(x) is all real numbers greater than or equal to zero (because you can never get a negative number when you square a number). SInce the range of g(x) is included in the domain of f(x), then the domain of f(g(x)) is all real numbers. Hope this wasn't too confusing.
How do you find the domain and range of composite functions?
The parameter methodIf you have each function given explicitly, say f(x) = x+3 and g(x)=x^2 and you want to find the domain and range of g(f(x)) then the easiest thing to do is form a function in one variable by passing the parameter through. So g(f(x))=g(x+3)=(x+3)^2. Finding the domain and range of this is simple, it's just a normal function. This method applies for higher orders of composition too. However, if you don't have the functions explicitly (we’re not always that lucky) or want to think about it in terms of sets read below.The set methodLet's say our composite function is g(f(x)). The easiest way to determine domain and range is to split the functions up.Start by taking f(x). Determine the range of that function. Take g(x) and determine it's domain. Do this as if they were individual functions - this should be fairly easy.Now take the intersection of those two sets, call it X. This is the set that links the two functions, it is mapable from the domain of g(f(x)) by carrying out f, and its mapable to the range of g(f(x)) by carrying out g.The domain of g(f(x)) is the preimage of X in f, or if you prefer f^-1(X)=Y. The range of g(f(x)) is g(X)=Z.This forms the sequence of equivalences g(f(Y))=g(X)=Z. So Y is the domain and Z is the range. This should be clear by f: Y → X and g: X → Z.This process can be extended to more highly composite functions such as f(g(h(x))) but requires a bit more moving backwards and forwards. I would start by determining the domain and range of g(h(x)) as above. Then carry out the process of intersecting the domain of f with the range of g(h(x)), call it X. You should then be fairly easily be able to get the total range by calculating f(X)=W. You'll then need to work back to determine the final domain of g(h(x)) that is allowable. To do this you carry out the preimage of g on X, I.e. g^-1(X) = Y. Then carry out the preimage of h on Y, I.e h^-1(Y) = Z. This gives the final domain.Written more nicely we will have found h: Z → Y, g: Y → X and f: X → W where Z is the domain and W is the range of f(g(h(x))).I hope this explanation helps. The set methodology requires a bit of thought but after some examples it becomes easier. If anyone has any examples they need help with feel free to comment or request me to answer another question and I'll get round to it.