f(x) = 2 - x - x^3
-
My first thought was that a cubic curve has two "bumps" in general, so in many cases a given value of f(x) will correspond to three values of x, which means it isn't possible to find an inverse without restricting the range.
However, in this case f'(x) = -1 - 3x² < 0 for all x
therefore f(x) is monotonically decreasing, so any given value of f(x) corresponds to only one value of x.
Hence the function has an inverse.
We can draw the graph of the inverse function:
It's the graph x = 2 - y - y³
but I don't know how to solve this equation to make y the subject,
which I expect is why you posted the question,
so I haven't been much use!!
But now I've gone to the wolframalpha website
http://www.wolframalpha.com
and typed in
inverse function of f(x)=2-x-x^3
and clicked on the = button
and it came up with a complicated-looking expression which I can't re-produce here. But if you go to that website yourself you can see it.
It also shows the graph of y = f(x) in a purplish colour,
and y = f⁻¹(x) in blue/ black (I can't tell which.)
EDIT: Yes, that's it, what eborg has typed out. But it looks much better on wolframalpha, so I suggest you look there.
However, in this case f'(x) = -1 - 3x² < 0 for all x
therefore f(x) is monotonically decreasing, so any given value of f(x) corresponds to only one value of x.
Hence the function has an inverse.
We can draw the graph of the inverse function:
It's the graph x = 2 - y - y³
but I don't know how to solve this equation to make y the subject,
which I expect is why you posted the question,
so I haven't been much use!!
But now I've gone to the wolframalpha website
http://www.wolframalpha.com
and typed in
inverse function of f(x)=2-x-x^3
and clicked on the = button
and it came up with a complicated-looking expression which I can't re-produce here. But if you go to that website yourself you can see it.
It also shows the graph of y = f(x) in a purplish colour,
and y = f⁻¹(x) in blue/ black (I can't tell which.)
EDIT: Yes, that's it, what eborg has typed out. But it looks much better on wolframalpha, so I suggest you look there.
-
(sqrt(3) sqrt(27 y^2-108 y+112)-9 y+18)^(1/3)/(2^(1/3) 3^(2/3))-(2/3)^(1/3)/(sqrt(3) sqrt(27 y^2-108 y+112)-9 y+18)^(1/3)