Show that $x=-ln(x)$,$x\in ]0,+\infty[$ has a unique solution.
Show that the iterative method $x_{n+1}=-ln(x_n)$ diverge giving $x_0 >0$
And purpose a method of approximating the solution.
My try:
Let $f(x)=x+ln(x)$ then $f'(x)=1+\frac 1x >0$ Hence $f$ is an increasing function and let $0<a<1$ and $b\geq 1$ so $f(a)f(b)<0$ by MVT we deduce there exists a unique solution on $]0,+\infty[$.
The second question let $g(x)=-ln(x)$ then $g'(x)=-\frac 1x$ Hence $|g'(x)|>0$ that is $g$ is not a contaction.
I suggest the Newton-Raphson method s.t $x_0>0,\quad x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$
Is this correct?