1. Compute the derivative of \( f(x) \). The function is \( f(x) = 12x^{4/3} - 6x^{1/3} \). Differentiating with respect to \( x \), we get \( f'(x) = 12 \cdot \frac{4}{3}x^{1/3} - 6 \cdot \frac{1}{3}x^{-2/3} \). Simplifying yields \( f'(x) = 16x^{1/3} - 2x^{-2/3} \).
2. Identify critical points by setting \( f'(x) = 0 \): \( 16x^{1/3} - 2x^{-2/3} = 0 \). Factorizing gives \( 2x^{-2/3}(8x - 1) = 0 \). This implies \( x^{-2/3} = 0 \) (invalid for \( x eq 0 \)) or \( 8x - 1 = 0 \). Solving \( 8x - 1 = 0 \) yields \( x = \frac{1}{8} \).
3. Evaluate \( f(x) \) at the critical point and endpoints. The critical point is \( x = \frac{1}{8} \). We evaluate \( f(x) \) at \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{8} \).
- For \( x = 0 \): \( f(0) = 12(0)^{4/3} - 6(0)^{1/3} = 0 \).
- For \( x = 1 \): \( f(1) = 12(1)^{4/3} - 6(1)^{1/3} = 12 - 6 = 6 \).
- For \( x = \frac{1}{8} \): \( f\left(\frac{1}{8}\right) = 12\left(\frac{1}{8}\right)^{4/3} - 6\left(\frac{1}{8}\right)^{1/3} \).
We have \( \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} \) and \( \left(\frac{1}{8}\right)^{4/3} = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \).
Substituting these values: \( f\left(\frac{1}{8}\right) = 12 \cdot \frac{1}{16} - 6 \cdot \frac{1}{2} = \frac{3}{4} - 3 = -\frac{9}{4} \).
4. Determine maximum and minimum values.
- \( f(0) = 0 \)
- \( f(1) = 6 \)
- \( f\left(\frac{1}{8}\right) = -\frac{9}{4} \)
The absolute maximum value is \( f(1) = 6 \).
The absolute minimum value is \( f\left(\frac{1}{8}\right) = -\frac{9}{4} \).
Final Answer:
- Absolute maximum value: \( 6 \) at \( x = 1 \).
- Absolute minimum value: \( -\frac{9}{4} \) at \( x = \frac{1}{8} \).