Answer is: option1

\( \frac{64}{27} \)

Solution:

To find the critical points, we first compute the derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx} \left( x - 2x^{2/3} \right) \] \[ = 1 - \frac{4}{3} x^{-1/3} \] \[ = 1 - \frac{4}{3} \frac{1}{x^{1/3}} \]

Setting \( f'(x) = 0 \) to find critical points:

\[ 1 = \frac{4}{3} \frac{1}{x^{1/3}} \] \[ x^{1/3} = \frac{4}{3} \] \[ x = \left( \frac{4}{3} \right)^3 \] \[ x = \frac{64}{27} \]

Compute the second derivative:

\[ f''(x) = \frac{d}{dx} \left( 1 - \frac{4}{3} x^{-1/3} \right) \] \[ = \frac{4}{9} x^{-4/3} \]

Since \( f''(x) > 0 \) for \( x > 0 \), this confirms a local minimum at \( x = \frac{64}{27} \).