Answer the following questions and show your work Let f be a

Answer the following questions and show your work: Let f be a function f: Zrightarrow Z x Z such that f(n) = (2n, n + 3). Verify whether this function is 1-1 and whether it is onto. Let f be a function/: R rightarrow 1 such that f(x, y, z) = xyz. Verify whether this function is 1-1 and whether it is onto. Prove that the function f: HR - {2} rightarrow R - {5} defined by f(x) = is a bisection and find f^-1. Consider the functions f defined as f(x) = and g(x) = (3x,x^2). Find the formulas for composition function f o F and go f.

Solution

(a). (p, q) in Z x Z is the preimage of ( p/2, q -3). When p is an odd integer, p/2 is not an integer. Therefore, f : Z Z x Z is not onto.

(b). xyz in is the pre-image of ( x, y, z), ( -x , -y, z), ( -x, y, -z) and ( x, -y, -z) in R³.Therefore, f is not 1-1. Further, for every real number x, there exist real numbers x/y, y/x, x² such that f( x/y, y/x, x) = x if x, y 0. Also, the pre-image of 0 in R is( 0, 0, 0) in R³ .Therefore, every element x in R has a pre-image ( x/y, y/x, x) in R³. Therefore, f is onto.

(c). Let y = (5x + 1)/ ( x -2). Then ( x -2) y = 5x + 1 or, xy -2y = 5x + 1 or, xy -5x = 2y + 1 or, x ( y -5) = 2y + 1 or, x = (2y + 1)/(y -5). On interchanging x and y, we have f^-1 (x) = (2x + 1)/(x -5).Since x 5, every element x in R – {5} has a pre-image (2x + 1)/(x -5) in R – {2}. Therefore, f is onto. Also, for, every unique x in R – {5}, (2x + 1)/(x -5) in R – {2} is unique. Therefore f is also 1-1. Thus f is a bijection. Also, f^-1 (x) = (2x + 1)/(x -5).

(d) f: RR is defined as f(x) = 1/(x² + 1) and g: R R x R is defined by g (x) = (3x, x²). Then f o f (x) = f (f(x)) = f [1/(x² + 1) ] = 1/[1/(x² + 1)² +1] = (x² + 1)²/[ 1 + ( x² + 1)²] = ( x⁴ + 2x² + 1)/( x⁴ + 2x² + 2). Also, g o f (x) = g ( f( x) ) = g [ 1/( x² + 1) ] = ( 3/( x² + 1) , 1/( x² + 1)²