f(x)=x^3-6x^2+9x+1, [2,4]
f' = 3 x^2 -12 x + 9
look for zeros in domain
x^2 -4x + 3 = 0
(x-3)(x-1) = 0
x = 1 or x = 3
3 is in our piece of domain between 2 and 4
What is function at x = 3 ? Calculate it.
Then also calculate the function at x = 2 and x = 4
Find the local and absolute extreme values of the function on the given interval. f(x)=x^3-6x^2+9x+1, [2,4]?
1 answer