The real valued function f(x) = 12x^4/3 - 6x^1/3, x ∈ [-8, 8] has abs…

The correct answer is Option (4) → 204

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

Critical points are

Compute derivative:

$f'(x) = \frac{d}{dx}\left(12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}\right) = 12 \cdot \frac{4}{3}x^{\frac{1}{3}} - 6 \cdot \frac{1}{3}x^{-\frac{2}{3}}$

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

Set $f'(x) = 0$:

Also, note that f′(x) is undefined at x = 0. Therefore, x = 0 is also a critical point and must be checked.

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

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

$\Rightarrow 8x^{\frac{1}{3} + \frac{2}{3}} = 1 \Rightarrow 8x = 1 \Rightarrow x = \frac{1}{8}$

f(0) = 12(0)^(4/3) − 6(0)^(1/3) = 0

Evaluate $f(x)$ at endpoints and critical point

$f(-8) = 12(-8)^{\frac{4}{3}} - 6(-8)^{\frac{1}{3}} = 12(16) - 6(-2) = 192 + 12 = 204$

$f(8) = 12(8)^{\frac{4}{3}} - 6(8)^{\frac{1}{3}} = 12(16) - 6(2) = 192 - 12 = 180$

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

Maximum at

$f(-8) = 204$, $f(8) = 180$, $f\left(\frac{1}{8}\right) = -\frac{9}{4}$, f(0) = 0

Absolute maximum value = 204