Finding first approximation for Newton's method without graphing How does one find the interval that encloses the root for large valued function like without graphing:
$$x^5-x^2+x=20000$$
I know that graphing with a software will give you the interval easily, but how do I find the two consecutive integers that enclose the root?
 A: I'm not sure why you're looking for a bracketing interval. That doesn't necessarily help for Newton's method. Consider, for example,
$$f(x) = -8 x^3+12 x^2-2.$$
It's easy to show by direct computation that $x=1/2$ is a simple root of $f$ and that $[0,1]$ is a bracketing interval. Unfortunately, $0$ and $1$ are both poor starting points for Newton's method here, since $f'(0)=f'(1)=0$.
Newton's method requires a single starting point. For this particular example, I took a look and thought - maybe $x_0$ should be around $$\sqrt[4]{20000} = 10\times \sqrt[4]{2},$$ which is probably just a little bit bigger than $10$. So I tried $x_0=12$ and it worked.
Application of Newton's method to arbitrarily contrived examples like this will almost certainly require ad hoc methods. When working with some particular family of functions that arise from an actual application, however, it's quite common that you can find reasonable starting values for that application. For example, this answer describes how Newton's method can be used to solve an equation that arises in the context of fractal dimension.
A: A bound (probably also named after Cauchy) for positive roots tells us to count the number $m$ of negative coefficients in $0=p(x)=x^n+a_{n-1}x^{n-1}+..+a_1x+a_0$ and then compute
$$
R = \max\left\{\sqrt[n-k]{m|a_k|}\middle|0\le k<n\land a_k<0\right\}.
$$
This is an upper bound for the positive roots. Because for $x>R$ one has 
$$
p(x)\ge x^n-\sum_{a_k<0}|a_k|x^k
> x^n-\sum_{a_k<0}\frac{R^{n-k}}mx^k
>x^n-\frac{m}mx^n=0$$
In this case, $$0=p(x)=x^5-x^2+x-20000,$$ we get $m=2$ and thus 
$$
R=\sqrt[5]{4\cdot 10^4}=10\cdot \sqrt[5]{\tfrac25}<8.5.
$$

As $-p(-x)=x^5+x^2+x+20000$, there will be no negative roots.
A: Hint.- Since the degree is odd, necessarily there is at least a real root $x_0$. What you have to do is to locate two consecutive integers $n$ and $n+1$ such that $f(n)$ and $f(n+1)$ have distinct signes. This procedure can fails when $x_0$ is a double roots in which case $x_0$ should be zero of the derivative $4x^4-2x+1$ which is not the case in this example.
