# 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]#?

##### 1 Answer

The absolute minimum value is

#### Explanation:

Absolute extrema of a differentiable function could either occur at the endpoints of the interval or at any critical values in the interval.

A critical value occurs at

Differentiate the function:

#f'(x)=4/3x^(1/3)-1-1/3x^(-2/3)#

It will be easier to find when this function

#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

Finding where the numerator equals

#4x-3x^(2/3)-1=0#

#4x-1=3x^(2/3)#

Cubing both sides:

#(4x-1)^3=27x^2#

#64x^3-48x^2+12x-1=27x^2#

#64x^3-75x^2+12x-1=0#

Note that

To determine if the other two solutions are real, do the long division to find the resulting quadratic:

#(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

Test the endpoints and critical values in the original function:

#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