Answer to Question #347212 in Discrete Mathematics for Douglas

Question

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

Answer 1

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

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

$$f^{-1}\circ g^{-1}=\dfrac{1}{2}(-\dfrac{1}{3}x+\dfrac{5}{3})-\dfrac{3}{2}$$ $$=-\dfrac{1}{6}x-\dfrac{2}{3}$$