The absolute extrema of a function are the largest and smallest y-values of the function on a given domain. This domain may be given to us (as in this problem) or it might be the domain of the function itself. Even when we are given the domain, we must consider the domain of the function itself, in case it excludes any values of the domain we are given.
#f(x)# contains the exponent #1/3#, which is not an integer. Luckily, the domain of #p(x)=root3(x)# is #(-oo,oo)# so this fact is not an issue.
However, we still need to consider the fact that the denominator cannot equal zero. The denominator will equal zero when #x=+-(1/3)=+-(sqrt(3)/3)#. Neither of these values lie in the given domain of #[2,9]#.
So, we turn to finding the absolute extrema on #[2,9]#. Absolute extrema occur at endpoints of the domain or at local extrema, that is points where the function changes direction. Local extrema occur at critical points, which are points in the domain where the derivative equals #0# or does not exist. Thus, we must find the derivative. Using the quotient rule:
#f'(x)=((3x^2-1)*(1/3)(9x^(-2/3))-9x^(1/3)*6x)/(3x^2-1)^2#
#f'(x)=((3x^2-1)*3x^(-2/3)-54x^(4/3))/(3x^2-1)^2#
#f'(x)=(9x^(4/3)-3x^(-2/3)-54x^(4/3))/(3x^2-1)^2#
#f'(x)=(-45x^(4/3)-3x^(-2/3))/(3x^2-1)^2#
If we factor #-3x^(-2/3)# out of the numerator, we have:
#f'(x)=(-3(15x^2+1))/(x^(2/3)(3x^2-1)#
There are no values of #x# on #[2,9]# where #f'(x)# does not exist. There are also no values on #[2,9]# where #f'(x)=0#. Thus there are no critical points on the given domain.
Using the "candidates test," we find the values of #f(x)# at the endpoints. #f(2) =( 9*root3(2))/(3*4-1)#=#( 9*root3(2))/11#
#f(9)=( 9*root3(9))/(3*9-1)#=#( 9*root3(9))/26#
A quick check on our calculators shows that:
#( 9*root3(2))/11##=1.030844495. . . # (absolute maximum)
#( 9*root3(9))/26##=0.7200290 . . .# (absolute minimum)
