Application of Differentiation (Doesn't understand) It's given the cubic equation $x^3-12x-5=0$.  Show graphically that the iteration $x_{n+1}=\sqrt[3]{12x_n+5}$ should be used to find the most negative root and the positive root, and the iteration $x_{n+1}=\dfrac{x^3_n-5}{12}$ should be used to find the other root.
This is the solution given by book, 
$$x_{n+1}=\sqrt[3]{12x_n+5}$$
$$F'(x_n)=\frac{4}{\sqrt[3]{(12x_n+5)^2}}$$
$$F'(-3)=0.41$$
$$F'(-0.5)=4$$
$$F'(3)=0.34$$
I've a problem here. Where the -3,-0.5 and 3 come from?
 A: The idea to find approximations of roots of $f(x) = 0$ starts with finding an integer $a$ such that a root lies between $a$ and $(a + 1)$. To ensure that this is so we need to guess some integers $a$ such that $f(a)f(a + 1) < 0$ i.e. $f(a)$ and $f(a + 1)$ are of opposite signs. Then by Intermediate Value Theorem there is a root of $f(x)$ between $a$ and $(a + 1)$.
Here $f(x) = x^{3} - 12x - 5$ and clearly we can see that $$f(0) = -5, f(1) = -16, f(2) = -21, f(3) = -14, f(4) = 9$$ so that there is a root in the interval $[3, 4]$. Again checking the negative integers we have $$f(0) = -5, f(-1) = 6$$ so that there is a root in interval $[-1, 0]$. Going to further negative values of $x$ we have $$f(-2) = 9, f(-3) = 4, f(-4) = -21$$ so that the third root lies between in $[-4, -3]$.
Now the method of iteration used here recasts the equation $f(x) = 0$ in the form $x = \phi(x)$ and uses the iteration $x_{n + 1} = \phi(x_{n})$ with suitable starting point $x_{0}$. The suitable starting is normally chosen to be one of the end-points of the interval in which the root lies. Sometimes a midpoint of the interval is also chosen.
Next we need to choose the $\phi(x)$ such that $|\phi'(x)| < k < 1$ (for some fixed number $k$) near the points of iteration $x_{i}$. This is needed to guarantee that the iteration converges to the desired root.
The solution has chosen $\phi(x) = \sqrt[3]{12x + 5}$ for one of the roots namely the one lying in interval $[-4, -3]$ and therefore the derivative $\phi'(-3) = 0.41$ has been calculated. (Your notation uses $F$ instead of $\phi$.) Similarly for the root in interval $[3, 4]$ the value $\phi'(3) = 0.34$ is used. In both the cases we see that the value of $|\phi'(x)|$ is less than $1$ and hence the iterations works.
However for the third root in $[-1, 0]$ it appears that the initial point of iteration is chosen as mid point $-0.5$ of the interval and then $\phi'(-0.5) = 4$ which is much greater than $1$. Choosing initial point as $0$ (or $-1$) would also give $|\phi'(x)| > 1$ so that the form of iteration needs to be changed for the iteration to converge.
For the root in $[-1, 0]$ we can write the equation as $$x = \frac{x^{3} - 5}{12}$$ so that $\phi(x) = (x^{3} - 5)/12$ and $\phi'(x) = x^{2}/4$. Clearly in this case $|\phi'(x)| < 1$ for any value of $x \in [-1, 0]$ and the iteration will converge.
I hope you now understand how we need to choose the form of iteration as well as the starting value for the iteration chosen.
Note: You should also read the textbook to understand why the condition $|\phi'(x)| < k < 1$ is needed to guarantee the convergence of iteration $x_{n + 1} = \phi(x_{n})$ to a root of equation $x = \phi(x)$. It is not very difficult to understand and you can revert back for more details.
