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I've been stumped on this problem for hours and cannot figure out how to do it from tons of tutorials.

Please note: This is an intro to calculus, so we haven't learned derivatives or anything too complex.

Here's the question:

Let $f(x) = x^5 + x + 7$. Find the value of the inverse function at a point. $f^{-1}(1035) = $___?

I tried setting $f(x)$ as $y$.. and solving for $x$. Clearly that doesn't help lol. I've tried many different approaches and cannot figure out the answer. I used wolframalpha, my textbook, notes, examples, and tons of Google searches and nothing makes sense. Can someone please help? Thanks!!

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    $\begingroup$ Guess and check! There is no nice formula for solving a fifth degree equation (that's a theorem). Your function is increasing, so there is a unique solution to $f(x)=1035$. If you fool around, you will find it very fast. $\endgroup$ – André Nicolas Sep 10 '13 at 4:20
  • $\begingroup$ Oh no, André edited his comment as I wrote my answer, and now they're the same. All credit to André, the comment FGITW. $\endgroup$ – davidlowryduda Sep 10 '13 at 4:24
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$$1035-7=1028=1024+4=4^5+4$$ Therefore $f^{-1}(1035)=4$.

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HINT(s)

  1. $f$ is an increasing function.
  2. Since $f$ is increasing, you will be able to modify your guesses to close in on the answer quickly.
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    $\begingroup$ So it's literally a guess and check type ordeal? $\endgroup$ – Cozen Sep 10 '13 at 4:24
  • $\begingroup$ @Justin: Why not? $\endgroup$ – davidlowryduda Sep 10 '13 at 4:25
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    $\begingroup$ Since $f$ is rapidly increasing, this will converge quickly. Take $x_0=1035^{\frac 15}$, ignoring the other terms. Then $x_i=(x_{i-1}^5-x_{i-1}-7)^{\frac 15}$ $\endgroup$ – Ross Millikan Sep 10 '13 at 4:32
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In general, polynomials won't have an inverse. This one happens to have one, but it's not fun to express, as far as I know.

Since you only need to find the inverse at a particular number, not any $y$, just plug it in and rearrange until something looks nice: $x^5 + x + 7 = 1035$ means $x(x^4 + 1) = 1028$. The factors of $1028$ are a good place to start.

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  • $\begingroup$ Ah, that's a nice one too. $\endgroup$ – davidlowryduda Sep 10 '13 at 4:26

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