Find for which value of a parameter $k$ a function is injective, surjective, or bijective

Question

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?

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.