This question was on one of our tests.I couldn't solve it but I'm curious to know the answer: Find the inverse function of: $$f(x)=(\ln(x))^2-\ln(x).$$ I found they domain of definition and studied the characteristics of this function but couldn't know how to find it's expression.
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1Is this function one to one? – Mhenni Benghorbal Jul 24 '13 at 14:42
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Hint
Let $X=\ln(x)$ and solve the quadratic equation $X^2-X-y=0$.
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You have:
$$y = (ln(x))^2 - ln(x)$$ Put $ln(x) = t$:
$$y = t^2 - t$$
Now, modify the RHS to make it a perfect square (Add and subtract 1/4). $$y = t^2 - 2(\frac{1}{2})t + (\frac{1}{2})^2 - \frac{1}{4} $$ There is NO change in the equation. It is absolutely the same as before! (Cancel out the terms and check yourself! Adding and subtracting the same term is very useful sometimes.) $$\therefore y = (t - \frac{1}{2})^2 - \frac{1}{4}$$
Now, add 1/4 to both sides and take the square root to get $t$. :)
mikhailcazi
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Doing the last step (taking the square root), you end up with two possibilities — namely, $\pm\sqrt{}$. – Clement C. Jul 24 '13 at 17:43
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You will need to be careful — your function is not injective. More precisely, $f_{|(0,\sqrt e]}$ and $f_{|[\sqrt e, \infty)}$ are both bijections from their domain to $[-\frac{1}{4}, \infty)$, so you'll have two "possible inverse functions" for $f$, depending on which restriction you choose to invert; respectively
$$\begin{align*} g_1&\colon y\in[-\frac{1}{4}, \infty)\mapsto \exp{\frac{1-\sqrt{1+4y}}{2}}\in(0,\sqrt{e}] \\ g_2&\colon y\in[-\frac{1}{4}, \infty)\mapsto \exp{\frac{1+\sqrt{1+4y}}{2}}\in[\sqrt{e},\infty) \end{align*} $$
(solutions found with the method @Sami Ben Romdhane described — solving $X^2-X-y=0$, with $X=\ln x$)
Clement C.
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