Question #347212

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
2022-06-06T09:28:55-0400

a. Let f(x1)=f(x2).f(x_1)=f(x_2). It means that


2x1+3=2x2+32x_1 + 3=2x_2 + 32x1=2x22x_1=2x_2x1=x2x_1=x_2

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

Let y=2x+3,yR.y=2x+3,y\in \R. Then


x=y232x=\dfrac{y}{2}-\dfrac{3}{2}

We see that xR yR.x\in \R\ \forall y\in \R.

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


b. Let g(x1)=g(x2).g(x_1)=g(x_2). It means that


3x2+5=3x2+5-3x_2 + 5= -3x_2 + 53x1=3x2-3x_1=-3x_2x1=x2x_1=x_2

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

Let y=3x+5,yR.y=-3x+5,y\in \R. Then


x=y3+53x=-\dfrac{y}{3}+\dfrac{5}{3}

We see that xR yR.x\in \R\ \forall y\in \R.

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


c.


gf=3(2x+3)+5=6x4g\circ f=-3(2x+3)+5=-6x-4gf=6x4g\circ f=-6x-4



d.


f(x)=2x+3,xRf(x)=2x+3, x\in \Ry=2x+3y=2x+3

Change xx and yy

x=2y+3x=2y+3

Solve for yy

y=12x32y=\dfrac{1}{2}x-\dfrac{3}{2}

Then


f1(x)=12x32f^{-1}(x)=\dfrac{1}{2}x-\dfrac{3}{2}




g(x)=3x+5,xRg(x)=-3x+5, x\in \Ry=3x+5y=-3x+5

Change xx and yy

x=3y+5x=-3y+5

Solve for yy

y=13x+53y=-\dfrac{1}{3}x+\dfrac{5}{3}

Then


g1(x)=13x+53g^{-1}(x)=-\dfrac{1}{3}x+\dfrac{5}{3}



e.


gf=6x4g\circ f=-6x-4y=6x4y=-6x-4

Change xx and yy

x=6y4x=-6y-4

Solve for yy

y=16x23y=-\dfrac{1}{6}x-\dfrac{2}{3}

Then


(gf)1(x)=16x23(g\circ f)^{-1}(x)=-\dfrac{1}{6}x-\dfrac{2}{3}




f1g1=12(13x+53)32f^{-1}\circ g^{-1}=\dfrac{1}{2}(-\dfrac{1}{3}x+\dfrac{5}{3})-\dfrac{3}{2}=16x23=-\dfrac{1}{6}x-\dfrac{2}{3}

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