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

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    2015
In summary, the inverse of f(x) for the interval (-∞,0) is not a function due to failing the horizontal line test. However, for the interval (0,∞), the inverse of f(x) is a one-to-one function given by g(x) = ∛(x+√(x^2+4))/2. The domain and range of the inverse for this interval are both (0,∞). To verify that a function is the inverse of f(x) for a specific interval, you can use the composition of functions method. Finally, for the interval [1,∞), the inverse of f(x) is given by g(x) = ∛(x+
  • #1
Ackbach
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Here is this week's POTW:

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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 $f^{-1}([0,2])$.

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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!
 
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  • #2
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*}
 

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

1. What is the inverse of f(x) for the interval (-∞,0)?

For the interval (-∞,0), the inverse of f(x) is not a function because it fails the horizontal line test. This means that for any given x-value, there are multiple y-values that satisfy the equation. Therefore, the inverse of f(x) for this interval is not well-defined.

2. Is the inverse of f(x) a one-to-one function for the interval (0,∞)?

Yes, for the interval (0,∞), the inverse of f(x) is a one-to-one function. This means that for any given x-value, there is only one corresponding y-value that satisfies the equation. The inverse of f(x) for this interval is given by the function g(x) = ∛(x+√(x^2+4))/2.

3. What is the domain and range of the inverse of f(x) for the interval (0,∞)?

The domain of the inverse of f(x) for the interval (0,∞) is the same as the range of f(x), which is also (0,∞). The range of the inverse of f(x) is the same as the domain of f(x), which is also (0,∞).

4. How do you verify that a function is the inverse of f(x) for a specific interval?

To verify that a function g(x) is the inverse of f(x) for a specific interval, you can use the composition of functions method. This means that you plug in g(x) as the input for f(x) and simplify the resulting expression. If the resulting expression is equal to x, then g(x) is the inverse of f(x) for that interval.

5. What is the inverse of f(x) for the interval [1,∞)?

For the interval [1,∞), the inverse of f(x) is given by the function g(x) = ∛(x+√(x^2+4))/2. This is the same inverse function as for the interval (0,∞), but with a restriction on the domain. This is because the original function f(x) has a vertical asymptote at x=1, so the inverse function has to be restricted to [1,∞) to avoid any discontinuities.

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