Given f(x) = sqrt(x-1) and g(x) = (x)/(x+2), find f g(x) and g f(x), stating their domains.

Here are the solutions to the problems. The functions are given as: f: x sqrt(x-1), with domain x R, x 1. g: x (x)/(x+2), with domain x R, x -2. a) Find f g(x) and g f(x), stating their domains. Step 1: Find f g(x). f g(x) = f(g(x)) = f((x)/(x+2)) Substitute g(x) into f(x): f((x)/(x+2)) = sqrt((x)/(x+2) - 1) Simplify the expression inside the square root: (x)/(x+2) - 1 = (x - (x+2))/(x+2) = (x - x - 2)/(x+2) = (-2)/(x+2) So, f g(x) = sqrt((-2)/(x+2)) Step 2: Determine the domain of f g(x). For f g(x) to be defined: 1. x must be in the domain of g, so x -2. 2. g(x) must be in the domain of f, so g(x) 1. (x)/(x+2) 1 (x)/(x+2) - 1 0 (x - (x+2))/(x+2) 0 (-2)/(x+2) 0 For this inequality to hold, since the numerator is negative, the denominator must be negative (and cannot be zero). x+2 < 0 x < -2 3. The expression inside the square root must be non-negative: (-2)/(x+2) 0. This leads to the same condition x < -2. Combining these conditions, the domain of f g(x) is \x R x < -2\. f g(x) = sqrt((-2)/(x+2)) Domain of f g(x) = \x R x < -2\ Step 3: Find g f(x). g f(x) = g(f(x)) = g(sqrt(x-1)) Substitute f(x) into g(x): g(sqrt(x-1)) = sqrt(x-1)sqrt(x-1)+2 Step 4: Determine the domain of g f(x). For g f(x) to be defined: 1. x must be in the domain of f, so x 1. 2. f(x) must be in the domain of g, so f(x) -2. sqrt(x-1) -2. Since sqrt(x-1) 0 for x 1, this condition is always satisfied. 3. The denominator of g(f(x)) cannot be zero: sqrt(x-1)+2 0. Since sqrt(x-1) 0, sqrt(x-1)+2 2. Thus, the denominator is never zero. This condition is always satisfied for x 1. Combining these conditions, the domain of g f(x) is \x R x 1\. g f(x) = sqrt(x-1)sqrt(x-1)+2 Domain of g f(x) = \x R x 1\ b) Show that g is not surjective. Step 1: Understand surjectivity. A function g: X Y is surjective if for every element y in the codomain Y, there exists at least one element x in the domain X such that g(x) = y. For g(x) = (x)/(x+2), the domain is X = R \-2\ and the codomain is Y = R. Step 2: Attempt to find x in terms of y. Let y = g(x). y = (x)/(x+2) Multiply both sides by (x+2): y(x+2) = x yx + 2y = x Rearrange to solve for x: 2y = x - yx 2y = x(1 - y) Step 3: Identify a value in the codomain that has no preimage. If 1 - y = 0, i.e., y = 1, the equation becomes 2(1) = x(0), which simplifies to 2 = 0. This is a contradiction. This means that there is no value of x in the domain of g for which g(x) = 1. Since 1 is an element of the codomain R, but it has no corresponding x in the domain, the function g is not surjective. Shown (ii) A relation R is defined on the set A = \1, 2, 3, 4, 5\ by a R b a + b = 2n, n N. List all the equivalence classes of A under R. Step 1: Interpret the relation condition. The condition a+b = 2n, n N means that a+b must be an even positive integer. Since a, b \1, 2, 3, 4, 5\, the smallest sum is 1+1=2 and the largest is 5+5=10. All these sums are positive. Therefore, the condition simplifies to $a+