if $f(x) = x^5+ x^3 + x$, then $f^{-1}(3) = ?$

I TRIED to do it, and I got this answer : 3/91 . I don't know if it is correct or not?

I tried to do this work like this: $$ y = x^5+x^3+x\\ y = x(x^4+x^2+1)\\ y/(x^4+x^2+1) = x\\ y = x/(x^4+x^2+1)\\ f^{-1}(x)= x/(x^4+x^2+1)\\ f^{-1}(3)=3/91$$

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    $\begingroup$ What is $f(1)$? Also notice that $f'(x)\gt0$. $\endgroup$ – robjohn Feb 7 at 22:41
  • $\begingroup$ Your fourth step is wrong. You go from $\frac{y}{x^4+x^2+1} = x$ to $y = \frac{x}{x^4+x^2+1}$, i.e., moved the denominator from one side to the other; you would have had to multiply through by $x^4+x^2+1$, and that would just have given you back your second line. $\endgroup$ – Arturo Magidin Feb 7 at 22:50
  • $\begingroup$ @ArturoMagidin how would i do it then? $\endgroup$ – Sports Highlights Feb 7 at 22:53
  • $\begingroup$ P.S. You'll have a very hard time finding an analytic formula for $f^{-1}$. But you aren't supposed to find one. You can use calculus to verify that $f$ is one-to-one, and then eyeball a number that $f$ maps to $3$, to get the value of $f^{-1}(3)$. You don't need to figure out how to invert every value, just $3$. $\endgroup$ – Arturo Magidin Feb 7 at 22:54

Let $y=y(x)$ is an inverse function.

Thus, $y^5+y^3+y=3,$ which since $y^5+y^3+y$ increases, gives $y=1$ and $f^{-1}(3)=1.$

  • $\begingroup$ How are your $y$ and $f$ (or $f^{-1}$) related to $x$ and $f$ from the original post? $\endgroup$ – Wolfgang Kais Feb 8 at 10:31
  • $\begingroup$ I just used the definition of inversed functions. You can see it in the net or I'll come back to the home and I'll write this definition. $\endgroup$ – Michael Rozenberg Feb 8 at 10:42
  • $\begingroup$ Thank you. Maybe it's just $y=f^{-1}(3)$ and thus $3=f(f^{-1}(3))=f(y)=y^5+y^3+y$... I think that I missed that one step. $\endgroup$ – Wolfgang Kais Feb 8 at 12:02
  • $\begingroup$ @Wolfgang Kais Indeed, $f(f^{-1}(3))=3$ because for all $x$ from the domain of $f^{-1}$ we have $f(f^{-1}(x))=x$, but $f^{-1}(3)=f^{-1}(f(1))=1.$ $\endgroup$ – Michael Rozenberg Feb 8 at 12:29
  • $\begingroup$ Thanks, but that's clear. The step missing in my mind was that you replaced $x$ from the definition of $f(x)$ with $y=f^{-1}(3)$, the desired value. The confusing thing was that you were looking at the problem from the $y$-axis (assuming that we already have an inverse function and solving an equation concerning the image of $f^{-1}$) while all other answerers (including me) solved an equation of the form $f(x)=3$, looking at the problem from the $x$-axis. :-) $\endgroup$ – Wolfgang Kais Feb 8 at 13:38

$f(x)-3 = x^5+x^3+x-3 = (x-1)(x^4+x^3+2x^2+2x+3) = 0$ $\Leftrightarrow x=1 \quad \lor \quad x^4+x^3+2x^2+2x+3 = 0$

Since $f'(x) = 5x^4 + 3x^2 + 1 \ge 1 > 0$ for all $x \in \mathbb R$, $f$ is strictly monotone increasing, thus $x=1$ must be the only (real) solution.


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