f(x) is a polynomial with real coefficients. Hence for any real numbers x, ...
• f(x) is a real-valued function
• f(x) is continuous on any closed interval of real numbers, such as [4, 6]
• f(x) is differentiable on any open interval of real numbers, such as (4, 6)
The conclusion is that there exists some "c" such that f'(c) = (f(6) -f(4)/(6 - 4).
The slope of interest is
m = (f(6) -f(4))/(6 -4) = (-44 -(-44))/2 = 0
The slope f'(x) is -20 +4x. It will be zero where
0 = -20 +4x
20 = 4x
5 = x
So, f'(5) = 0 = m
The point "c" of interest is c = {5}
