Question #149168
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?
1
Expert's answer
2020-12-14T16:11:18-0500

a) What happens if we graph both f and f1f^{-1} on the same set of axes, using the x-axis for the input to both f and f1f^{-1}?


first of all f and f1f^{-1} are inverses of each other.

Let's see if we graph both f and f1f^{-1} on the same set of axes. Let's denote f1=gf^{-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=x3x^3{-2 < x < 2}, y=x1/3x^{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 f1f^{-1} on the same set of axes.


y=x3x^3 {-2 < x < 2}, y=x1/3x^{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 = x3x^3 and g =x1/3x^{1/3}

We find f(g(x)) = (x1/3)3=x(x^{1/3})^3=x and g(f(x)) = (x3)1/3(x^3)^{1/3} = x

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

  • let's find their domain and the range:

the domain of x3x^3 is (-\infty, +\infty ) and its range is (-\infty , +\infty ),

the domain of x1/3x^{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 =x2x^2 and y = x\sqrt{x} , y = - x\sqrt{x}





  • let g = x2x^2 and g =x1/2x^{1/2} and x1/2-x^{1/2}

we find f(g(x)) = (x1/2)2=x(x^{1/2})^2=x and g(f(x)) = (x2)1/2(x^2)^{1/2} = x

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

  • lets find their domain and range

the domain of x2x^2 is (-\infty , +\infty ) and the range is [0, +\infty )

x1/2-x^{1/2} -x^(1/2) and x1/2x^{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 f2f^{2} to represent f \circ f and fn+1=fnf^{n+1}=f^n \circ f.

Is it true that f2f^2 \circ f = f\circ f2f^2? Why? Is the set {g:R \rightarrow R l g\circ f=f o g} infinite? Why?

f2f^2 = f \circ f = x+1+1 = x+2

f2ff^2 \circ f = x+1+2 = x+3

ff2f \circ f^2 = x+2+1 = x+3, so yes they are the same;

f3f^3 =f2ff^2 \circ f = x+3

f4=f3ff^4=f^3 \circ f = x+4

...

Suppose fnf^n = x+n,

fn+1=fnff^{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


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