Let $$f(x)=\ln(x^4-x^2+1)-2\ln(x^2+1)+2\sqrt3\arctan(\frac{2x^2-1}{\sqrt3}) + kx -k\ln x;$$ $$f_{*}(x)=\ln(x^4-x^2+1)-2\ln(x^2+1)+2\sqrt3\arctan(\frac{2x^2-1}{\sqrt3}) + kx;$$ $$f_{**}(x)=\ln(x^4-x^2+1)-2\ln(x^2+1)+2\sqrt3\arctan(\frac{2x^2-1}{\sqrt3});$$ for $k \in \mathbb{R}$.

I am a little stumped: how can I determine for what values of $k$, $f$ is (1) injective, (2) surjective; (3) bijective; and for what value of $k$, $f_{*}$ is $(1^*)$ injective, $(2^*)$ surjective; $(3^*)$ bijective? I have never done similar exercises before, and I'd be grateful if you could show me how to do it.

Also, can you show me how to draw the qualiative plot of $f(x)$, $f_{*}$ and $f_{**}$? I know I have to study the first and second derivative, and so on, but in practice, I found the task difficult for $f_{**}$ and I don't know how to proceed with the other two functions, which have parameters. Could you show me how to proceed?

  • $\begingroup$ Sorry, these look like very boring and useless exercises. $\endgroup$
    – egreg
    Commented Mar 1, 2015 at 17:01
  • $\begingroup$ Prof. @egreg, I know, but I cannot solve them, and I would be very grateful if you could explain to me how to proceed. $\endgroup$ Commented Mar 1, 2015 at 22:29
  • $\begingroup$ Are you sure that the first term is $\ln(x^2-x^2+1)$? It probably is $\ln(x^2-x+1)$. Anyway, the drill is the same: for surjectivity find the limit at $-\infty$, the limit at $\infty$ (they must be distinct infinities); for injectivity, look at the derivative: the functions must be monotonic. $\endgroup$
    – egreg
    Commented Mar 1, 2015 at 22:33
  • $\begingroup$ Prof. @egreg, yes, there was a typo (the correct writing was $\ln(x^4-x^2+1)$; thank you. I've edited the question. If you've got some time, could you show me how to solve the problem in detail? $\endgroup$ Commented Mar 2, 2015 at 13:09

1 Answer 1


We have easily $$ \lim_{x\to-\infty}f_{**}(x)=\lim_{x\to\infty}f_{**}(x)=\pi\sqrt{3} $$ so $$ \lim_{x\to\infty}f(x)=\lim_{x\to\infty}f_{*}(x)= \begin{cases} \infty & \text{if $k>0$,}\\ \pi\sqrt{3} & \text{if $k=0$,}\\ -\infty & \text{if $k<0$,} \end{cases} $$ while $$ \lim_{x\to0}f(x)=\begin{cases} -\infty & \text{if $k>0$,}\\ -\pi\sqrt{3}/3 & \text{if $k=0$,}\\ \infty & \text{if $k<0$,} \end{cases} $$ and $$ \lim_{x\to-\infty}f_{*}(x)=\begin{cases} -\infty & \text{if $k>0$,}\\ \pi\sqrt{3} & \text{if $k=0$,}\\ \infty & \text{if $k<0$.} \end{cases} $$ Thus $f_{**}$ is neither injective nor surjective.

Thus surjectivity is ruled out for $f$ and $f_{*}$ when $k=0$. Also injectivity of $f_{*}$ is ruled out for $k=0$. On the other hand, $f$ and $f_{*}$ are surjective for $k\ne0$.

Now compute the derivatives and try deciding whether the functions $f$ and $f_{*}$ are monotonic.


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