I have been asked by a friend to solve the following inequality. She is in her first year of a Physics university course. I am in my third of a Mathematics, so I don't know what this says about my skills... But here I am, asking for help.
$$ (3x-1)\ln x + x -1 \leq 0$$
Here is what I got:
If I call the left side $f(x)$, it is a $C^\infty$ function over $\mathbb{R}^+$. Its second derivative is always positive, so the function is convex. Therefore the solution of the above inequality must be a (closed) interval. Obviously $f(1)=0$, so $1$ is one end of said interval, but I can't find the other.
To note: proving that the solution is a closed interval is rather easy. The exercise specifically asks to find said interval, and implies that it can be done in an exact, analytic way (withut numerical approximations).
Yes, we have $$((3x-1)\ln{x}+x-1)''=\frac{3x+1}{x^2}>0,$$ which says that the equation $$(3x-1)\ln{x}+x-1=0$$ has two roots maximum.
$1$ is a root and the second is $0.1754...$ and we got the answer: $$(0.1754...,1).$$ We can not write an exact value of the second root in the elementary functions.