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]?

May 29, 2016

The absolute minimum value is $- 1$, located at $x = 1$, and the absolute maximum value is $5 \sqrt{6} - 6$, 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 $f \left(c\right)$ is defined and $f ' \left(c\right) = 0$ or is undefined.

Differentiate the function:

$f ' \left(x\right) = \frac{4}{3} {x}^{\frac{1}{3}} - 1 - \frac{1}{3} {x}^{- \frac{2}{3}}$

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 $\frac{3 {x}^{\frac{2}{3}}}{3 {x}^{\frac{2}{3}}}$.

$f ' \left(x\right) = \frac{\frac{4}{3} {x}^{\frac{1}{3}} - 1 - \frac{1}{3} {x}^{- \frac{2}{3}}}{1} \left(\frac{3 {x}^{\frac{2}{3}}}{3 {x}^{\frac{2}{3}}}\right)$

$f ' \left(x\right) = \frac{4 x - 3 {x}^{\frac{2}{3}} - 1}{3 {x}^{\frac{2}{3}}}$

A critical value exists at $x = 0$, since this is where $f ' \left(x\right)$ is undefined because the denominator, $3 {x}^{\frac{2}{3}}$, is equal to $0$.

Finding where the numerator equals $0$, which is when $f ' \left(x\right) = 0$, is a little trickier:

$4 x - 3 {x}^{\frac{2}{3}} - 1 = 0$

$4 x - 1 = 3 {x}^{\frac{2}{3}}$

Cubing both sides:

${\left(4 x - 1\right)}^{3} = 27 {x}^{2}$

$64 {x}^{3} - 48 {x}^{2} + 12 x - 1 = 27 {x}^{2}$

$64 {x}^{3} - 75 {x}^{2} + 12 x - 1 = 0$

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:

$\frac{64 {x}^{3} - 75 {x}^{2} + 12 x - 1}{x - 1} = 64 {x}^{2} - 11 x + 1$

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:

$f \left(- 1\right) = {\left(- 1\right)}^{\frac{4}{3}} - \left(- 1\right) - {\left(- 1\right)}^{\frac{1}{3}} = 1 + 1 + 1 = 3$

$f \left(0\right) = {0}^{\frac{4}{3}} - 0 - {0}^{\frac{1}{3}} = 0$

$f \left(1\right) = {1}^{\frac{4}{3}} - 1 - {1}^{\frac{1}{3}} = 1 - 1 - 1 = - 1$

$f \left(6\right) = {6}^{\frac{4}{3}} - 6 - {6}^{\frac{1}{3}} = {6}^{\frac{1}{3}} \left(6 - 1\right) - 6 = 5 \sqrt{6} - 6 \approx 3.0856$

From here we see that the absolute minimum value is $- 1$, located at $x = 1$, and the absolute maximum value is $5 \sqrt{6} - 6$, located at $x = 6$.