approximate the root of perturbed polynomial 
Approximate the root of $$f(x)=(x-1)(x-1)(x-3)(x-4)-10^{-6}x^6$$ near $r=4$.

Do I have to use iterative method of finding the root, such as Bisection, Secant, etc? Is there other way?
 A: If
$f(x)
=(x-1)(x-2)(x-3)(x-4)-10^{-6}x^6
$,
if $c$ is small,
$\begin{array}\\
f(4+c)
&=(3+c)(2+c)(1+c)(c)-10^{-6}(4+c)^6\\
&=6c(1+c/2)(1+c/3)(1+c)-10^{-6}4^6(1+c/4)^6\\
&\approx 6c(1+c/2+c/3+c)-(2/5)^6(1+6c/4)\\
&=6c(1+11c/6)-(2/5)^6(1+3c/2)\\
&=c(6+(3/2)(2/5)^6) +11c^2-(2/5)^6\\
&\approx 6c -(2/5)^6\\
\end{array}
$
If this is zero,
$c = \frac{(2/5)^6}{6}
\approx 0.0006826
$
so the root is about
$4.0006826$,
which agrees very nicely
with Moo's much more accurate answer.
A: If you want an approximation, the implicit function theorem is a useful tool.
Let $\phi(x,\epsilon) = (x-1)(x-2)(x-3)(x-4)-\epsilon x^6$. It is not too
difficult to compute 
${\partial \phi(4,0) \over \partial x} = 6$, ${\partial \phi(4,0) \over \partial \epsilon} = -4^6$. Hence there is a function $\xi$ defined in a neighbourhood of $\epsilon=0$ such that $\phi(\xi(\epsilon), \epsilon) = 0$,
and
${\partial \xi(0) \over \partial \epsilon} = - ({\partial \phi(4,0) \over \partial x})^{-1} {\partial \phi(4,0) \over \partial \epsilon} = - {-4^6 \over 6} = {4^6 \over 6}$.
Hence we expect $\xi({1 \over 10^6}) \approx \xi(0) + {\partial \xi(0) \over \partial \epsilon} {1 \over 10^6}=4 +{4^6 \over 6} {1 \over 10^6} \approx 4.000683$.
