The real valued function $f(x) = 12x^\frac{4}{3} - 6x^\frac{1}{3}, x \in [-8, 8]$ has absolute maximum value equal to

Given: $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$ on the interval $x \in [-8, 8]$

To find the absolute maximum value, we need to check critical points (where the derivative $= 0$ or doesn't exist) and the endpoints of the interval.

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Find the derivative:

$f'(x) = 12 \cdot \frac{4}{3}x^{\frac{4}{3}-1} - 6 \cdot \frac{1}{3}x^{\frac{1}{3}-1}$

$f'(x) = 16x^{\frac{1}{3}} - 2x^{-\frac{2}{3}}$

Rewriting:

$f'(x) = 16x^{\frac{1}{3}} - \dfrac{2}{x^{\frac{2}{3}}}$

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Find critical points by setting $f'(x) = 0$:

$16x^{\frac{1}{3}} - \dfrac{2}{x^{\frac{2}{3}}} = 0$

Multiply both sides by $x^{\frac{2}{3}}$:

$16x^{\frac{1}{3}} \cdot x^{\frac{2}{3}} - 2 = 0$

$16x - 2 = 0$

$x = \dfrac{1}{8}$

Also, $f'(x)$ doesn't exist at $x = 0$ (due to $x^{-\frac{2}{3}}$ in denominator).

Critical points: $x = 0$ and $x = \dfrac{1}{8}$

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Evaluate the function at critical points and endpoints:

At $x = -8$:

$f(-8) = 12(-8)^{\frac{4}{3}} - 6(-8)^{\frac{1}{3}}$

Since $(-8)^{\frac{1}{3}} = -2$:

$f(-8) = 12(-2)^4 - 6(-2) = 12(16) + 12 = 192 + 12 = 204$

At $x = 0$:

$f(0) = 12(0)^{\frac{4}{3}} - 6(0)^{\frac{1}{3}} = 0$

At $x = \dfrac{1}{8}$:

$f\left(\dfrac{1}{8}\right) = 12\left(\dfrac{1}{8}\right)^{\frac{4}{3}} - 6\left(\dfrac{1}{8}\right)^{\frac{1}{3}}$

Since $\left(\dfrac{1}{8}\right)^{\frac{1}{3}} = \dfrac{1}{2}$:

$f\left(\dfrac{1}{8}\right) = 12\left(\dfrac{1}{2}\right)^4 - 6\left(\dfrac{1}{2}\right) = 12 \cdot \dfrac{1}{16} - 3 = \dfrac{3}{4} - 3 = -\dfrac{9}{4}$

At $x = 8$:

$f(8) = 12(8)^{\frac{4}{3}} - 6(8)^{\frac{1}{3}}$

Since $(8)^{\frac{1}{3}} = 2$:

$f(8) = 12(2)^4 - 6(2) = 12(16) - 12 = 192 - 12 = 180$

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Comparing all values:

$f(-8) = 204$

$f(0) = 0$

$f\left(\dfrac{1}{8}\right) = -\dfrac{9}{4}$

$f(8) = 180$

The absolute maximum value is $204$