Question

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]`?

Answer 1

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

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 `x=c` if `f(c)` is defined and `f'(c)=0` or is undefined.

Differentiate 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 both sides:

`(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.

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 `0` and `1`.

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