MATH116 Lecture Notes - Lecture 16: Entire Function

• School: University of Waterloo
• Department: Mathematics
• Course: MATH116
• Professor: Fiona Dunbar

Document Summary

Tutorial solutions 1: let f (x) = 3x + 1 and g(x) = x 1. Find the following: (a) f (g(1)) (b) g(f (x)) (c) f (f (x)) (b) g(f (x)) =(cid:112)f (x) 1 = Solution: x2 2 + 2 = x2 = |x| (note this is not necessarily x itself). The domain of g (the inside function) is all of r, so we don"t get any restrictions beyond what we see in the simpli ed answer. The domain of f (b) g f (x) = g(f (x)) = (f (x))2 2 = ( is restricted, due to the square root. Although we later square the result and the expression simpli es, the entire function is still g applied to f (x), so f (x) must be de ned too. This means the range will be exactly the same as the domain (every y value is equal to the corresponding x value!).