Study of a function to draw its graphic

Question

I'm studying the following function: $$f(x)=x+\frac{ax}{\ln x}$$ So I divided the work in two part for $a>0$ and for $a<0$. I started from the first and studying it I found that it is defined for $(0,1)$ and $(1,{+\infty})$, then I know that it is positive in $x\to(0,e^{-a})$ negative in $x\to(e^{-a},1)$ and then positive again. Studying the limits in $0$ and $1$, I find for the first $0^+$ and for the second $+\infty$ from the right side of $1$ and the $-\infty$ from the left side. And finally the study of its derivative. Here I'm stuck. I find maximum and minimum in places that are not acceptable referring to the previous info that I collected (for example a maximum in a place in which the function is not defined).. Where is my mistake? Can you show me how to deal with this kind of function with a parameter? My problem is especially in the study of the first derivative where clearly I have problem in "interpreting" the influence of $a$. And possibly can you give me hints in studying the second case ($a<0$)? Thank you in advance!

Answer 1

Various thoughts that might help the sketch:

• There is a third possibility of $a=0$, though this leads to the simple $f(x)=x$

• $f(x)-x \to 0$ as $x \to \infty$

• You have $f(e^{-a})=0$

• $f'(x)$ is close to $1$ for very small positive $x$

• Looking at $f'(x)=0$:

  • there are no real solutions for $\log_e(x)$ when $-4 \lt a \lt 0$, suggesting no local maxima or minima for positive $x$;
  • when $a=-4$, $f'(e^2)=0$ but $f'(x)$ is positive around $x=e^2$, so this is probably neither a minimum or a maximum
  • when $a \lt - 4$ there is both a local maximum and a local minimum in $(1,e^{-a})$
  • when $a \gt 0$ there is a local maximum in $(0,e^{-a})$ and a local minimum in $(1,\infty)$