Lagrange interpolation: Evaluation of error in interpolation

Question

I was given the following nodes:
$x_1=0$
$x_2=1$
$x_3=8$
$x_4=27$
and was asked to interpolate the Lagrange polynomial of the function $f(x)=\sqrt[3]{x}$ (meaning, I have the values: $(0,0)$,$(1,1)$,$(8,2)$,$(27,3)$ and need to interpolate the Lagrange polynomial), and to evaluate the error of interpolation at the given interval $[0,27]$.
So, I constructed the Lagrange polynomials;
$L_0=-\frac{1}{216}(x-1)(x-8)(x-27)$
$L_1=\frac{1}{182}x(x-8)(x-27)$
$L_2=-\frac{1}{1064}x(x-1)(x-27)$
$L_3=\frac{1}{4446}x(x-1)(x-8)$
and found the interpolating polynomial, $P(x)$ using the formula:
$$P(x)=\sum_{i=0}^{n=3}f_i\cdot L_i(x)$$ So far everything seems good (I hope).
The problem arises when I try to evaluate the error...
Using the following formula:
$$E(x)=f(x)-P(x)=\frac{MN}{(n+1)!}$$ Where:
$$M=\max_{x\in[0,27]}|f^{(4)}(x)|$$ $$N=\max_{x\in[0,27]}|x(x-1)(x-8)(x-27)|$$ When I try to evaluate $M$, I get: $f^{(4)}(x)=\frac{10}{27}x^{-\frac{8}{3}}=\frac{10}{27\sqrt[3]{x^8}}$, but this function has no maximum value at $[0,27]$!
How could that be?

Answer 1

First, your "error formula" is wrong: $MN/(n+1)!$ is an upper-bound of the absolute value of $E(x)$, not its value.

The objective of the exercise you're doing is to show you that this upper-bound can be undefined, while you get an actual error $\max_x E(x)$ which is finite. Thus you should calculate $\max_x E(x)$.