Question

Find the absolute maximum and minimum values of the function: \[ f(x) = 12x^{4/3} - 6x^{1/3}, \quad x \in [0, 1]. \]

Hint:

To find absolute extrema of a function on a closed interval, evaluate the function at critical points and endpoints, and compare the values.

Answer 1

1. Find the derivative of \( f(x) \): The given function is: \[ f(x) = 12x^{4/3} - 6x^{1/3}. \] Differentiate with respect to \( x \): \[ f'(x) = 12 \cdot \frac{4}{3}x^{1/3} - 6 \cdot \frac{1}{3}x^{-2/3}. \] Simplify: \[ f'(x) = 16x^{1/3} - 2x^{-2/3}. \] 2. Critical points: To find critical points, set \( f'(x) = 0 \): \[ 16x^{1/3} - 2x^{-2/3} = 0. \] Factorize: \[ 2x^{-2/3}(8x - 1) = 0. \] This gives: \[ x^{-2/3} = 0 \quad \text{(not valid as \( x \neq 0 \))}, \quad \text{or} \quad 8x - 1 = 0. \] Solve \( 8x - 1 = 0 \): \[ x = \frac{1}{8}. \] 3. Evaluate \( f(x) \) at critical points and endpoints: The critical point is \( x = \frac{1}{8} \). Evaluate \( f(x) \) at \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{8} \): - At \( x = 0 \): \[ f(0) = 12(0)^{4/3} - 6(0)^{1/3} = 0. \] - At \( x = 1 \): \[ f(1) = 12(1)^{4/3} - 6(1)^{1/3} = 12 - 6 = 6. \] - At \( x = \frac{1}{8} \): \[ f\left(\frac{1}{8}\right) = 12\left(\frac{1}{8}\right)^{4/3} - 6\left(\frac{1}{8}\right)^{1/3}. \] Simplify: \[ \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2}, \quad \left(\frac{1}{8}\right)^{4/3} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}. \] Substitute: \[ f\left(\frac{1}{8}\right) = 12 \cdot \frac{1}{16} - 6 \cdot \frac{1}{2} = \frac{3}{4} - 3 = -\frac{9}{4}. \] 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 \( x = \frac{1}{8} \).