Question 897313
Better if you are studying Calculus & Analytical Geometry 1.


{{{df/dx=3x^2-12x+9=0}}}


{{{x^2-4x+3=0}}}
{{{x=(4+- sqrt(16-4*3))/2}}}
{{{x=(4+- 2)/2}}}
{{{highlight_green(x=1)}}}  or  {{{highlight_green(x=3)}}}
Those roots for x are where the extreme values of f will occur.


You are told to find the extreme value on the interval [1,2] and that is where x=1 is.
The x=3 is outside of that interval.
This extreme value at x=1 is {{{highlight(f(1))=1^3-6*1^1+9*1+8=1-6+9+8=-5+17=highlight(12)}}}