MHB What is the inverse of the function f(x)=x^3-3x for different intervals?

  • Thread starter Thread starter Ackbach
  • Start date Start date
  • Tags Tags
    2015
Click For Summary
The discussion focuses on finding the inverse of the function f(x) = x^3 - 3x over specified intervals. Participants are tasked with calculating f^{-1}([-2,2]), f^{-1}((2,18)), f^{-1}([2,18)), and f^{-1}([0,2]). The original poster notes that no one answered the problem correctly and provides their own solution. The thread emphasizes understanding the behavior of the cubic function and its implications for finding inverses in different ranges. Overall, the discussion highlights the complexities involved in determining function inverses for specific intervals.
Ackbach
Gold Member
MHB
Messages
4,148
Reaction score
94
Here is this week's POTW:

-----

Let $f:\mathbb{R}\to\mathbb{R}$ be the function given by $f(x)=x^3-3x$. Calculate $f^{-1}([-2,2]), f^{-1}((2,18)), f^{-1}([2,18)),$ and $f^{-1}([0,2])$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
No one answered this week's POTW correctly. Here is my solution:

A plot of the function follows using Desmos. You can adjust it so that it captures all $y$-values from $-2$ to $18$.
[desmos="-10,10,-10,10"]x^3-3*x[/desmos]
From this graph, we can see by inspection that
\begin{align*}
f^{-1}([-2,2])&=[-2,2] \\
f^{-1}((2,18))&=(2,3) \\
f^{-1}([2,18))&=\{-1\}\cup [2,3) \\
f^{-1}([0,2])&=[-\sqrt{3},0]\cup[\sqrt{3},2].
\end{align*}
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K