To use Newton's method on interval $[a,b]$ we need to guarantee that

1. $f(a)f(b)<0$ on the interval which is true for $[0,1]$.
2. $f'(x)$ and $f''(x)$ are continuous on the interval $[a,b]$ (which they are)
3. $f'(x) \neq 0$ on $[a,b]$ and $f''(x)$ does not change signs on $[a,b]$

Then if $f(x\_{0})f''(x\_{0})>0$ the method will converge.

I have the following exercise in my textbook and I don't understand a part of the solution:

Using Newton's method and starting approximation $x\_{0} = 1$ find the root of the equation $$f(x)=e^{x}-\sin(x)-1.5$$
Given solution:
$$f'(x) = e^{x}-\cos(x)$$
$$f''(x) = e^{x}+\sin(x)$$
$f'(x)$ and $f''(x)$ are both positive on the given interval, we can therefore use it.
We verify that $f(0)f''(0)>0$ and then start the algorithm itself.

My questions for this part of the solution:

1. How can I determine what are the function values for a function like $f'(x)$ or $f''(x)$? Is it possible to graph them by hand?
2. $f'(x)=0$ for $x=0$ which seems to be in contradiction with the convergence criteria. Is the solution right?

Thanks.