Inverse functions

  • Thread starter PFuser1232
  • Start date
  • #1
479
20
Is there a way to formally prove that if ##f## and ##g## are multiplicative inverses of each other, then ##f^{-1} (x) = g^{-1} (\frac{1}{x})##?
 

Answers and Replies

  • #2
Dr. Courtney
Education Advisor
Insights Author
Gold Member
2020 Award
3,289
2,447
I'd start by testing a number of special cases and see what approach the trickier special cases suggest.

Simple special cases almost never suggest approaches to proof, but the tricky special cases often do.
 
  • #3
pasmith
Homework Helper
1,918
557
Is there a way to formally prove that if ##f## and ##g## are multiplicative inverses of each other, then ##f^{-1} (x) = g^{-1} (\frac{1}{x})##?
Let [itex]h[/itex] be the function which takes [itex]x[/itex] to [itex]1/x[/itex]. Now if [itex]f(x)g(x) = 1[/itex] for all [itex]x[/itex] then [itex]f = h \circ g[/itex]. Then [itex]f^{-1} = g^{-1} \circ h^{-1}[/itex]. But [itex]h = h^{-1}[/itex] so [itex]f^{-1} = g^{-1} \circ h[/itex] as required.
 
  • Like
Likes PFuser1232
  • #4
479
20
Let [itex]h[/itex] be the function which takes [itex]x[/itex] to [itex]1/x[/itex]. Now if [itex]f(x)g(x) = 1[/itex] for all [itex]x[/itex] then [itex]f = h \circ g[/itex]. Then [itex]f^{-1} = g^{-1} \circ h^{-1}[/itex]. But [itex]h = h^{-1}[/itex] so [itex]f^{-1} = g^{-1} \circ h[/itex] as required.
Perfect. Thanks!
 

Related Threads on Inverse functions

  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
1
Views
775
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
4
Views
687
  • Last Post
Replies
6
Views
2K
Top