Answer to Question #149168 in Algebra for Patrick

Question

What happens if we graph both f and f^{-1} on the same set of axes, using the x-axis for the input to both f and f^{-1} ?

[Suggestion: go to www.desmos.com/calculator and type y=x^3 {-2 < x < 2}, y=x^{1/3} {–2 < x < 2}, and y = x {–2 < x < 2}, and describe the relationship between the three curves.] Then post your own example discussing the difficulty of graph both f and f^{-1} on the same set of axes.

Suppose f:R \rightarrow R is a function from the set of real numbers to the same set with f(x)=x+1. We write f^{2} to represent f \circ f and f^{n+1}=f^n \circ f. Is it true that f^2 \circ f = f \circ f^2? Why? Is the set {g:R \rightarrow R l g \circ f=f \circ g} infinite? Why?

Answer 1

a) What happens if we graph both f and $f^{-1}$ on the same set of axes, using the x-axis for the input to both f and $f^{-1}$?

first of all f and $f^{-1}$ are inverses of each other.

Let's see if we graph both f and $f^{-1}$ on the same set of axes. Let's denote $f^{-1}=g$

• (f _o g)(x) = x , x in the domain of g and (g _o f)(x) = x , x in the domain of f
• the domain of f is equal to the range of g and the range of f is equal to the domain of g.
• their graphs are symmetrical reflections of each other with respect to the line y = x.

b ) [Suggestion: go to www.desmos.com/calculator and type y=$x^3${-2 < x < 2}, y=$x^{1/3}$ {–2 < x < 2}, and y = x {–2 < x < 2}, and describe the relationship between the three curves.] Then post your own example discussing the difficulty of graph both f and $f^{-1}$ on the same set of axes.

y=$x^3$ {-2 < x < 2}, y=$x^{1/3}$ {–2 < x < 2}, and y = x {–2 < x < 2}. Let's see their graphs on www.desmos.com/calculator.

The following properties can be seen clearly from the graph

• let g = $x^3$ and g =$x^{1/3}$

We find f(g(x)) = $(x^{1/3})^3=x$ and g(f(x)) = $(x^3)^{1/3}$ = x

f(g(x)) = g(f(x))

• let's find their domain and the range:

the domain of $x^3$ is (-$\infty$, +$\infty$ ) and its range is (-$\infty$ , +$\infty$ ),

the domain of $x^{1/3}$ is (-$\infty$ , +$\infty$ ) and its range is (-$\infty$ , +$\infty$ )

it can be seen the domain of f is equal to the range of g and the range of f is equal to the domain of g.

• the functions are a reflection of each other with respect to the graph of y = x.

my own example

y =$x^2$ and y = $\sqrt{x}$ , y = - $\sqrt{x}$

• let g = $x^2$ and g =$x^{1/2}$ and $-x^{1/2}$

we find f(g(x)) = $(x^{1/2})^2=x$ and g(f(x)) = $(x^2)^{1/2}$ = x

f(g(x)) = g(f(x))

• lets find their domain and range

the domain of $x^2$ is (-$\infty$ , +$\infty$ ) and the range is [0, +$\infty$ )

$-x^{1/2}$ -x^(1/2) and $x^{1/2}$ have the domain [0, +$\infty$ ) and the range (-$\infty$ , +$\infty$ )

it can be seen the domain of f is equal to the range of g and the range of f is equal to the domain of g

• the functions are a reflection of each other with respect to the graph of y = x.

c) Suppose f:R $\rightarrow$ R is a function from the set of real numbers to the same set with f(x)=x+1. We write $f^{2}$ to represent f $\circ$ f and $f^{n+1}=f^n$ $\circ$ f.

Is it true that $f^2$ $\circ$ f = f$\circ$ $f^2$? Why? Is the set {g:R $\rightarrow$ R l g$\circ$ f=f o g} infinite? Why?

$f^2$ = f $\circ$ f = x+1+1 = x+2

$f^2 \circ f$ = x+1+2 = x+3

$f \circ f^2$ = x+2+1 = x+3, so yes they are the same;

$f^3$ =$f^2 \circ f$ = x+3

$f^4=f^3 \circ f$ = x+4

...

Suppose $f^n$ = x+n,

$f^{n+1}=f^n \circ f$ = x+1+n = x+(n+1)

so the pattern holds by induction,

F $\circ$ Finv = x so when you graph them on

the same plot, they shall be SYMMETRIC about

the line y=x