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1. notthejake Says:
   March 25th, 2008 at 7:29 am

   1) you can't take the square root of a negative number, so 16 – x >= 0, or x <= 16 (-inifinity,16] in interval notation

   2) because f(g) is sqrt(16 – g) you can replace x with g in the domain above:
   g<= 16
   but we know g = x^4, so x^4 <= 16
   take fourth root of both sides, and remember the inequality:
   -2 <= x <= 2 (or [-2,2])

   the domain of g is all real numbers, so that doesn't limit the domain any

   c) domain of g is all real numbers.
   since g(f) must be defined, that means f must be defined, and the domain of this composite is the same as the domain of f: x <= 16 (-infinity, 16]

2. Scranderberry Says:
   March 25th, 2008 at 7:29 am

   The domain is the x values for which the function exists.

   f(x) = sqrt(16 – x)
   this works only when 16-x is positive so 16-x>0
   x<16 (x is an element of the reals)

   g(x) = x^4.

   fog = f(x^4)=sqrt(16 – x^4)
   which works when 16-x^4 >0
   x^4<16
   -2<x<2 (x is an element of the reals)

   gof=g(sqrt(16 – x))=sqrt(16 – x)^4
   =(16-x)^2
   this is do-able for all values of x

   or x is any real

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