These are detailed solutions to the questions on functions.

Solution to Question 1: f(0) is found by substituting x by 0 in the formula f(x) = 2x + 4 since 0 is less than 2. f(0) = 4 f(2) is found by substituting x by 2 in the formula f(x) = 2x + 4 since 2 is less than or equal to 2. f(2) = 2*2 + 4 = 8 f(4) is found by substituting x by 4 in the formula f(x) = 2x - 1 since 4 is greater than 2. f(4) = 2*4 - 1 = 7

Solution to Question 2: To answer the question, we need to solve the given equation for y. y = + or - √ (x^{ 2} +2) y is not a function of x because for one value of the independent variable x (input) we obtain two values for the dependent variable y (output).

Solution to Question 3: The expression inside the square root is positive for any real value of x; therefore the domain of function f is the set of all real numbers.

Solution to Question 4: Evaluate f(-9) by substituting x by -9 in the formula of the function. f(-9) = |-9 - 1| + 3 = 10 + 3 = 13

Solution to Question 5: (f _{o} g)(6) is calculated as follows (f _{o} g)(6) = f(g(6)) Let us first evaluate g(6) g(6) = √ (6 + 3) = 3 We now substitute g(6) by 3 in f(g(6)) f(g(6)) = f(3) Finally evaluate f(3) f(3) = 2×3 + 2 = 8 and (f _{o} g)(6) = 8.

Solution to Question 6: f(x) = 0 gives the following equation (x^{ 2} + 2 x - 3) / (x - 1) = 0 The denominator is not equal to 0 for values of x not equal to 1. The above equation leads to (x^{ 2} + 2 x - 3) = 0 Solve the above equation by factoring (x - 1)(x + 3) = 0 The above equation has two solutions x = 1 and x = -3. x = 1 is not a solution to f(x) = 0 because It is a value of x that makes the denominator of f(x) equal to zero. So the only value of x for which f(x) = 0 is -3.

Solution to Question 7: g(x) = g(x) leads to an equation. 3x + √(x) = 2x + 6 Rewrite the equation as follows √(x) = - x + 6 Square both sides and simplify x = (-x + 6)^{ 2} x = x ^{ 2} + 36 - 12 x Rewrite in standard form. x ^{ 2} - 13 x + 36 = 0 Solve the above quadratic equation. x = 4 and x = 9 Since we squared both sides of the equation, extraneous solutions may be introduced but can be eliminated by checking. After checking, x = 4 is the only value of x that makes f(x) = g(x).

Solution to Question 8: Solve the given equation for y. y = -1 - (x^{ 2} + 2) or y = -1 + (x^{ 2} + 2) The two solutions mean that for one value of x we obtain two values of y and y is not a function of x.

Solution to Question 9: The domain of function f is the set of all real values of x for which f(x) is real. |x - 1| is always positive except when x = 1 which makes it equal to 0. But |x - 1| is in the denominator. Therefore the domain of f is the set of all real numbers except 0.