Answer to Question #343244 in Discrete Mathematics for bkay

Question #343244

Let the function f : R → R and g : R → R be defined by f(x) 2x + 3 and g(x) = -3x + 5.

a. Show that f is one-to-one and onto.

b. Show that g is one-to-one and onto.

c. Determine the composition function g o f

d. Determine the inverse functions f -1 and g -1 .

e. Determine the inverse function (g o f) -1 of g o f and the composite f -1 o g -1 . 


1
Expert's answer
2022-05-22T23:59:30-0400

a. Let "f(x_1)=f(x_2)." It means that


"2x_1 + 3=2x_2 + 3"

"2x_1=2x_2"

"x_1=x_2"

The function "f(x)=2x+3" is one-to-one from "\\R" to "\\R."

Let "y=2x+3,y\\in \\R." Then


"x=\\dfrac{y}{2}-\\dfrac{3}{2}"

We see that "x\\in \\R\\ \\forall y\\in \\R."

The function "f(x)=2x+3" is onto from "\\R" to "\\R."


b. Let "g(x_1)=g(x_2)." It means that


"-3x_2 + 5= -3x_2 + 5"

"-3x_1=-3x_2"

"x_1=x_2"

The function "g(x)=-3x+5" is one-to-one from "\\R" to "\\R."

Let "y=-3x+5,y\\in \\R." Then


"x=-\\dfrac{y}{3}+\\dfrac{5}{3}"

We see that "x\\in \\R\\ \\forall y\\in \\R."

The function "g(x)=-3x+5" is onto from "\\R" to "\\R."


c.


"g\\circ f=-3(2x+3)+5=-6x-4"

"g\\circ f=-6x-4"



d.


"f(x)=2x+3, x\\in \\R"

"y=2x+3"

Change "x" and "y"


"x=2y+3"

Solve for "y"


"y=\\dfrac{1}{2}x-\\dfrac{3}{2}"

Then


"f^{-1}(x)=\\dfrac{1}{2}x-\\dfrac{3}{2}"


"g(x)=-3x+5, x\\in \\R"

"y=-3x+5"

Change "x" and "y"


"x=-3y+5"

Solve for "y"


"y=-\\dfrac{1}{3}x+\\dfrac{5}{3}"

Then


"g^{-1}(x)=-\\dfrac{1}{3}x+\\dfrac{5}{3}"



e.


"g\\circ f=-6x-4"

"y=-6x-4"

Change "x" and "y"

"x=-6y-4"

Solve for "y"

"y=-\\dfrac{1}{6}x-\\dfrac{2}{3}"

Then


"(g\\circ f)^{-1}(x)=-\\dfrac{1}{6}x-\\dfrac{2}{3}"




"f^{-1}\\circ g^{-1}=\\dfrac{1}{2}(-\\dfrac{1}{3}x+\\dfrac{5}{3})-\\dfrac{3}{2}"

"=-\\dfrac{1}{6}x-\\dfrac{2}{3}"


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