Inverse of function $f(x)=x^3+3x+1$

Question

If $f:R \to R, f(x)=x^3+3x+1$ then find the inverse of $f(x)$.

$f(x)$ is one to one (as it is increasing function for $x \in R$) and onto as well (Range is same as co-domain) but I don't know how to find inverse here as I am not able to write $x$ as a function of y.

Could someone help me with this?

In this case Cardano's method will work, https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method — Will Jagy, May 19 '16 at 17:48
Write $y=f(x)$ to get a cubic equation in $x$. Then just solve it for $x$ and you will get an expression involving $y$ as $y$ appears in the coefficient. — joy, May 19 '16 at 17:50
Possible duplicate of Finding the inverse of $f(x) = x^3 + x$ — Travis Willse, May 19 '16 at 18:14

score 1 · Answer 1 · answered May 19 '16 at 17:58

1

For many — in fact, I would argue for most — purposes, the best expression for the inverse is

$g$ is the inverse of $f$

or maybe

$g(x) = a$ where $a$ is the unique value satisfying $f(a) = x$

and work with it in those terms. For example, do you want the derivative of $g$? The best way to compute it is with implicit differentiation:

$$f(g(x)) = x \implies g'(x) f'(g(x)) = 1 \implies g'(x) = \frac{1}{f'(g(x))} $$

Even if you had an algebraic formula for $g$, you would probably strongly prefer to compute the derivative in this fashion.

Another example: want the asymptotic behavior of $g(x)$ as $x \to \infty$? Use the behavior of $f$ and a change of variable:

$$ 1 = \lim_{x \to \infty} \frac{f(x)}{x^3} = \lim_{y \to \infty} \frac{f(g(y))}{g(y)^3} = \lim_{y \to \infty} \frac{y}{g(y)^3} $$

from which we conclude

$$ \lim_{x \to \infty} \frac{g(x)}{x^{1/3}} = 1 $$

answered May 19 '16 at 17:58

Sorry but I still don't get how this information would help me in finding in inverse of concerned function. – Akira May 19 '16 at 18:03
1

@Akira: The inverse is trivial to find: it's simply $f^{-1}$. The point of my post is an attempt to disabuse you of the notion that $f^{-1}$ is not an adequate formula for whatever purpose you have for the inverse. – May 19 '16 at 18:13
In total agreement with this answer, I would say that although the existence of the Cardano formula is of great theoretical interest, it’s unlikely that any understanding would be gained by its use, for instance in further computations. On the other hand, dealing with $f^{-1}$ as an increasing function, continuous and even infinitely differentiable, will best be done simply by referring to it in the way that the answerer here recommends. – Lubin May 19 '16 at 18:46

score 0 · Answer 2 · answered May 19 '16 at 18:21

write $x^3 + 3x + 1 = a,$ so $x^3 + 3x + (1-a) = 0.$ There is exactly one real solution $x.$ Therefore Cardano's method will allow us to find this $x$ explicitly with real-valued square roots and cube roots. That is, we avoid https://en.wikipedia.org/wiki/Casus_irreducibilis regardless of the value of $a.$ https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Inverse of function $f(x)=x^3+3x+1$

2 Answers2