Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 1

You need to evaluate the local absolute extrema of the function, hence, you need to find the zeroes of the equation f'(x) = 0.
You need to evaluate the derivative such that:
f'(x) = 3x^2 - 12x + 9
You need to solve for x the equation f'(x) =0:
3x^2 - 12x + 9= 0 => x^2 - 4x + 3 = 0
x_(1,2) = (4+-sqrt(16-12))/2
x_(1,2) = (4+-2)/2
x_1 = 3 ; x_2 = 1
You need to evaluate the function at critical points:
f(1) = 1^3 - 6*1^2 + 9*1 + 1
f(1) = 1 - 6 + 9 + 1
f(1) = 5
f(3) = 3^3 - 6*3^2 + 9*3 + 1
f(3) = 27 - 54 + 27 + 1
f(3) = 1
You need to evaluate the function at the end points of interval:
f(2) = 2^3 - 6*2^2 + 9*2 + 1 = > f(2) =8 - 24 + 18 + 1 = 3
f(4) = 4^3 - 6*4^2 + 9*4 + 1 => f(4) = 64 - 96 + 36 + 1= -3
Hence, the absolute maximum of the function, on the interval [2,4], is 5 and it occurs at x = 1 and the absolute minimum of the function is -3 and it occurs at x = 4.