Find roots of ln equation $$f(x)=\frac{\ln x}{x}-(x-1)$$
When I derivative the function, I got this:
$\frac{1-\ln x-x^2}{x^2} $
And my problem is I don't know how to find the roots or zero of $1-\ln x-x^2$ to continue my exercise, but I know that the roots of it is $x=1$ because the exercise told us to show that at that point the graph got the peak, but I don't know how to find it.
Sorry for any language mistakes, English is not my first language, and thanks for the help in advance.
 A: $$f(x)=\frac{\ln x}{x}-(x-1)$$
I assume the domain of the function to be $x>0$, since $\ln x$ is not defined for $x<0$ and $\frac{\ln x}{x}$ is not defined for $x=0$
Then $$f'(x) =\frac{x\cdot \frac{1}{x} -\ln x}{x^2}-1 = \frac{1-\ln x-x^2}{x^2}$$
Now notice that $f'(x)<0\quad \forall x>1$ which implies $f(x)$ is a decreasing function for $x>1$
Also notice that for $0<x<1 \quad \ln x<0$
Therefore $1>\ln x +x^2$ for $0<x<1$. Therefore for $0<x<1,f'(x)>0$
So we have that $0<x<1, f'(x)>0$ and $x>1, f'(x)<0$
Therefore at $x=1$ we must have a maximum, and we have shown above that it is a global maximum. 
So the maximum of $f(x)$ is given by $f(1) = 0$
A: $$g(x)=1-\ln x - x^2\to g'(x)=-\left(\frac{1}{x}+2x\right)<0\text{ for any $x>0$}.$$
But we also know that 


*

*$g(1/e)=2-1/e^2>0$;

*$g(1)=0$

*$g$ is decreasing because $g'(x)<0$.


It means that $x=1$ is the only one roots of $g(x).$
Without using a numerical method you can't find the roots. The question in your book gave you a hint about the root, so like I just did, you can show that it is unique.
Now you just have to show if it is a maximum or a minimum which is not so difficult.
