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How to invert this function? $$ y = e^{\arctan(x^5)} $$

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I mean how to inverse this function, I'm sorry. – TomDavies92 Oct 24 '12 at 15:22
HINT: Exchange x and y and re-arrange. Also note that the domain of the inverse will be restricted due to the trig function. – Epictetus Oct 24 '12 at 15:26
Four answers and I'm the only one to up-vote this question. (Actually, the reason I up-voted it is that it's a nice opportunity to say "What gets done last gets undone first.", which I think is the best way to think about this sort of thing.) – Michael Hardy Oct 24 '12 at 17:45
A point of grammar: "inverse" is a noun; "invert" is a verb. The English language is rather chaotic about things like this. – Michael Hardy Oct 25 '12 at 18:22
up vote 6 down vote accepted

$\newcommand{\leftlong}{\longleftarrow\!\shortmid}$ What gets done last gets undone first: $$ \begin{array}{rcccccl} x & \longmapsto & x^5 & \longmapsto & \arctan(x^5) & \longmapsto & \exp(\arctan(x^5)) = y \\[12pt] \sqrt[5]{\tan(\log_e y)} & \leftlong & \tan(\log_e y) & \leftlong & \log_e y & \leftlong & y \end{array} $$

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I guess by "solve", you mean "find the inverse $x=f(y)$".

$$y=e^{\arctan(x^{5})}\Leftrightarrow \log{y}=\arctan(x^{5})\Leftrightarrow \tan{(\log{y})}=x^{5}\Leftrightarrow(\tan(\log{y}))^{1/5}=x.$$

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Your last two $\iff$s don't hold. They concern non-injective functions, so the "inverse" will need to pick arbitrary preimages. – Lord_Farin Oct 24 '12 at 15:24
you forgot the last step i.e. switch x and y so the answer is $y = (\tan(\ln(x)))^{\frac{1}{5}}$ – Yash Jain Oct 24 '12 at 15:26
very helpful! Now I got it, thanks! – TomDavies92 Oct 24 '12 at 15:30
@Lord_Farin The last biconditional holds perfectly well (there is always a fifth root of $x \in \mathbb{R}$), and it's reasonable to assume that the student understands that $\tan(x)$ has to be restricted to a suitable interval. (Edit: Whoops, I thought this was recent.) – Chris Nov 1 '12 at 22:34

$${}x=\sqrt[5]{\tan(\log y)}$$

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