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$$x^3+8x+3=y$$ How am I suppose to make $y$ the subject? I'm not sure how to find its inverse, for example how to I find $f^{-1}(12)$

Thomas Andrews
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    $$1^3+8\times 1+3 = 12$$ – hamam_Abdallah Apr 29 '21 at 23:43
  • if $x=1$ then $y=12$ There is a formula for the inverse function using Cardano's method, but it is intricate – Will Jagy Apr 29 '21 at 23:43
  • The only way to find a general $f^{-1}(y)$ is to use the cubic formula, which is messy. For particular values of $y,$ however, you might just find the $x$ by trial and error. https://math.vanderbilt.edu/schectex/courses/cubic/ – Thomas Andrews Apr 29 '21 at 23:48

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Consistent with Will Jagy's comment, since $\displaystyle (4 \times 8^3) + (27 \times 3^2) > 0,$
you are guaranteed that the equation only has 1 real root, which may be computed as shown by Cardano's formula.

Also, if $\displaystyle f(x)=x^3 + 8x + 3$, then $\displaystyle f′(x)=3x^2 + 8.$

This means that $f(x)$ is strictly increasing.

This means that $f(x)$ is injective, which implies that there is a general function $g(y)$ such that $f(x) = y \iff g(y)=x.$

Therefore, the general inverse function that you speak of does exist.

The general function $g(y)$ may be analytically constructed as follows:

  • Make $y$ a variable in $\Bbb{R}.$

  • Construct $f_y(x)=(x^3 +8x + 3) − y.$

  • Then, $f_y(x) = 0 \iff f(x) = y.$

Therefore, following the link in in the first paragraph of my answer, simply apply Cardano's formula to $f_y(x)$ to obtain the value of $x,$ in terms of $y$. This construction, is in effect a function $g(y)$ such that $g(y)=x.$

Notice that with $f(x)$ strictly increasing and injective, $f_y(x)$ is simply $f(x)$ plus a constant. This means that the behavior of $f_y(x)$ will parallel $f(x)$, and so $f_y(x)$ will also only have 1 real root.

user2661923
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