Find the absolute maximum value and the absolute minimum value of #f(x) = x^(4/3) −x−x^(1/3)# on the interval# [−1, 6]#?

Question

Paisley Johnson · Accepted Answer

The absolute minimum value is #-1#, located at #x=1#, and the absolute maximum value is #5root3 6-6#, located at #x=6#.

A differentiable function's absolute extrema may appear at any critical value within the interval or at the interval's endpoints. A critical value occurs at #x=c# if #f(c)# is defined and #f'(c)=0# or is undefined. Distinguish the function: #f'(x)=4/3x^(1/3)-1-1/3x^(-2/3)# It will be easier to find when this function #=0# or is undefined by writing it as a fraction. Do so by multiplying the function by #(3x^(2/3))/(3x^(2/3))#. #f'(x)=(4/3x^(1/3)-1-1/3x^(-2/3))/1((3x^(2/3))/(3x^(2/3)))# #f'(x)=(4x-3x^(2/3)-1)/(3x^(2/3))# A critical value exists at #x=0#, since this is where #f'(x)# is undefined because the denominator, #3x^(2/3)#, is equal to #0#. Finding where the numerator equals #0#, which is when #f'(x)=0#, is a little trickier: #4x-3x^(2/3)-1=0# #4x-1=3x^(2/3)# Cubing each side separately: #(4x-1)^3=27x^2# #64x^3-48x^2+12x-1=27x^2# #64x^3-75x^2+12x-1=0# Note that #64-75+12-1=0#, which means that #x=1# is a solution to this cubic. Perform the long division to find the resulting quadratic in order to ascertain the veracity of the other two solutions: #(64x^3-75x^2+12x-1)/(x-1)=64x^2-11x+1# Here, the quadratic's discriminant reveals that the remaining two solutions are imaginary, so we are left with the two critical values of #0# and #1#. Examine the original function's endpoints and critical values: #f(-1)=(-1)^(4/3)-(-1)-(-1)^(1/3)=1+1+1=3# #f(0)=0^(4/3)-0-0^(1/3)=0# #f(1)=1^(4/3)-1-1^(1/3)=1-1-1=-1# #f(6)=6^(4/3)-6-6^(1/3)=6^(1/3)(6-1)-6=5root3 6-6approx3.0856# From here we see that the absolute minimum value is #-1#, located at #x=1#, and the absolute maximum value is #5root3 6-6#, located at #x=6#.

Emilia Aguilar · Answer

undefined The absolute maximum value of f(x) = x^(4/3) −x−x^(1/3) on the interval [−1, 6] is approximately 4.732, and the absolute minimum value is approximately -1.732.