What is the maximum of $w(x)=\prod\limits_{i=0}^8(x-x_i)$ on the interval $[-1,1]$, with
$\bullet$ equidistant nodes $x_i$, $(x_0=-1,x_8=1).$
$\bullet$Chebyshev nodes, $\displaystyle x_i=\cos\Big(\frac{2i+1}{2(n+1)}\pi\Big)$
I can insert the values, differentiate, set equal to zero and find the possible extrema, but is there another, faster way to do it, I don't think, that this is the goal of the exercise, thanks in advance.
The derivative is $$\sum_{j=0}^8 \prod\limits_{i=0, i\ne j}^8(x-x_i)$$ and the extrema must lie between the nodes. The function $w(x)$ alternates sign $+,-,+,-,\dots$ on each sub-interval
In the case of the Chebyshev points, the function can be defined as a sinusoid (the 9th Chebyshev polynomial): $A \cos(9\arccos(x))$ - so its max is the amplitude $A$ - and the maxima must happen at $9\arccos(x^*) = 2 k \pi$: then $A = \prod\limits_{i=0}^8(x^*-x_i)$ , for example we could use $x^* = \cos(2\pi/9)$