Exercise: Let $f(x)=-4x^{2}+e^{-x} +4x$.
Show that in $[0, ∞)$ it has only one root $ρ$.
Consider the sequence $(x_{n})$, $n ≥ 0$, that results if we apply Newton's method to approach the root of $f$ at $[0, ∞)$. Find out for which values $x_{0} ≥ 0$, the sequence $(x_{n})$ converges to the root $ρ$.
I used Bolzano's theorem to see that there is a root between $0$ and $2$
I found $f'(x)=-8x-e^{-x}+4$. So I can imagine that $f$ is increasing from $x=0$ to $x=k, k<1$. And the root is not in this area. $f$ is decreasing after $x=k$ and so there is only one root in $(k, ∞)$. That root is close to $1$.
That's my best try to prove the first argument. I can't even see the second one. I tried to write Newton's method but none of the Theorems I know can't help me figure it out.
Partial answer...
The derivative $f'$ has a root $x^* \approx 0.418$. If $x_0 < x^*$, Newton's method will always produce negative values and will not converge to the root of interest in this case.
For large values of $x_0$, since $g(x) = x-\frac{f(x)}{f'(x)} \sim \frac 14 + \frac x2$, Newton's iterations will decrease by a factor of $1/2$ and eventually fall in the interval $[1,2]$. You can check convergence in this interval...
If $x_0 \in (x^*, 1)$, $x_1 >1$ and we fall in the previous case.
So, finally, Newton's method will converge to this particular root if $x_0 > x^*$.
Note: When $x_0 < x^*$ Newton's method also converges, but to one of the negative roots of $f$.