The real valued function \(f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}, x \in [-8, 8]\) has absolute maximum value equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 204
On a closed interval, the absolute maximum is the largest of the values of f at the critical points and at the end points.
\( \displaystyle f'(x) = 16x^{\frac{1}{3}} - 2x^{-\frac{2}{3}} = \frac{2(8x - 1)}{x^{\frac{2}{3}}} \)
So \(f'(x) = 0\) at \(x = \frac{1}{8}\), and \(f'(x)\) is not defined at \(x = 0\). Both are critical points.
\(f(-8) = 12(-2)^4 - 6(-2) = 192 + 12 = 204\)
\(f(0) = 0\)
\( \displaystyle f\left(\frac{1}{8}\right) = 12 \cdot \frac{1}{16} - 6 \cdot \frac{1}{2} = \frac{3}{4} - 3 = -\frac{9}{4} \)
\(f(8) = 12(2)^4 - 6(2) = 192 - 12 = 180\)
The largest value is 204, at \(x = -8\).
Why the other options are incorrect
Option (1) 0 — Incorrect. This is \(f(0)\), the value at a critical point, not the largest value.
Option (2) \(-\displaystyle\frac{9}{4}\) — Incorrect. This is \(f\left(\frac{1}{8}\right)\), the absolute minimum value.
Option (3) 180 — Incorrect. This is \(f(8)\), the value at the other end point; \(f(-8) = 204\) is larger.