I need some tips, because I'm stuck. I have been given the function
$g(x)=11x+6-\cos(x)$
The text provides me with the given information:
There exists a solution $c$ in the interval $(-1,0)$ (This is not necessary to show). Use Newton's method and show with an initial value $x_0=0$ that $$c<\cdots<x_{n+1}<x_n<\cdots<x_1<x_0=0$$ (concavity should be used in your argument)
What I'm thinking: I made a sketch of the function, and I see that the inequality is right, but I have a hard time explaining why. The next task asks me to calculate both $x_1$ and $x_2$, so I don't think I should use those numbers to prove the inequality. So Newton's method gives us
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ and so,
$$x_{n+1}= x_n - \frac{11 x_n +6-\cos(x_n)}{11+\sin(x_n)}$$
But I don't see how I should use those facts. I really appreciate some tips.
Probably the easiest way to explain this short of actually doing it is to explain an analogous problem: show that Newton's method applied to $f(x)=x^2-2$, for any $x_0>\sqrt{2}$, has the property $\sqrt{2} < \dots < x_{n+1} < x_n < \dots < x_0$.
Another way of saying this is that for all $n \geq 0$, $\sqrt{2}<x_{n+1}<x_n$.
So now let $n \geq 0$. Start with the lower bound. For induction, assume $x_n>\sqrt{2}$, then conclude that $f(x_{n+1})>0$, since a strictly convex function lies strictly above its tangent lines and the tangent line from $x_n$ vanishes at $x_{n+1}$. Noting that $x_{n+1}>0$ by some algebra, you conclude $x_{n+1}>\sqrt{2}$.
By induction, now you know $x_n>\sqrt{2}$ for all $n$. Then you get the upper bound by using the lower bound to conclude that $f(x_n)/f'(x_n)>0$ for all $n$.