Find roots of ln equation

Question

$$f(x)=\frac{\ln x}{x}-(x-1)$$

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.

Answer 1

$$f(x)=\frac{\ln x}{x}-(x-1)$$

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 $f(1) = 0$

Answer 2

$$g(x)=1-\ln x - x^2\to g'(x)=-\left(\frac{1}{x}+2x\right)<0\text{ for any $x>0$}.$$

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.