a) Prove algebraically that f: R → R, defined by f(x) = 3x is a 1-1, onto (bijective) map. Note that we have to prove that i) f(x1) = f (x2) → x1 = x2 b) For each y ER, we can find x E R such that f(x) = y

Let f1(x)=3x and f2(x)=|x|. Graph both functions on the interval 2x2. Show that these functions are linearly dependent in the vector space C[0,1], but linearly independent in C[1,1].

Select the following functions that are bijective. f3: R → R, f(x) = 14 f₂: R → R, f₂(x) 2 f₁:R → R, f₁(x) = x + x + x + x + 4 f4: R → R, f4(x) = -{ = x x², if x 20 X, if x <0

Show that f : ℝ → ℝ, f(x) = 5x − 3 is a bijective function.

f: X→ Y and g: Y→V be bijective functions. Show the following: ) gof: X → V is injective. ) gof: X → V is surjective. ) (gof)-¹=f-¹og=¹ Use: v injective if x1 ‡xn → v(x1) = v(xn)

The function f is defined as follows: f: R² → R² with f(x, y) = (2x, x+y) Which statement is correct? f is NOT a linear map and it is not one to one. f is a linear map but it is not injective. f is a linear map and it is injective. f is a NOT a linear map but it is one to one.

a) Give an example of an algebraic function f : Z → Z that is injective but not surjective. You do not need to defend your answer b) Let A = {a, b} and B = {1,2}. Give an element of A x B × A.

Show that the function f : R – {3} → R – {5} defined by f(x) = *, is bijective and determine f-(x) for x eR – {5}.

f: X→ Y and g: Y→V be bijective functions. Show the following: ) gof: X → V is injective. ) gof: X→ V is surjective. ) (gof)-¹ = f¹og-¹ v injective if x1 #xn → v(x1) = v(xn)

a function. tion. (b) Let R= {(x,y) E R × R : 1 < x < 4 or a = 6}, and S = {(x,y) ER × R : –1 < y < 1}. i. Find the relations S1 and R-1 o S. ii. Represent the relations R, S, RUS and RnS in the (r, y)- plane.