Answer is: option3

Solution:

Finding Critical Points

Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined.

1. Where \( f'(x) = 0 \):

\[ \frac{3\sin(2x)}{x^2} = 0 \] The fraction is zero when the numerator is zero: \[ \sin(2x) = 0 \] Solving for \( x \): \[ 2x = k\pi, \quad k \in \mathbb{Z} \] \[ x = \frac{k\pi}{2} \] Within the interval \( (0,10) \), we find the values of \( x \): \[ \frac{k\pi}{2} < 10 \] Approximating \( \pi \approx 3.14 \): \[ k < \frac{10 \times 2}{3.14} \approx 6.37 \] The integer values of \( k \) are \( k = 1,2,3,4,5,6 \), corresponding to: \[ x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi \] This gives six critical points.

Where \( f'(x) \) is Undefined:

The denominator \( x^2 \) is undefined at \( x = 0 \), but \( 0 \notin (0,10) \), so it does not contribute to additional critical points.

The function \( f \) has six critical values in \( (0,10) \).

Answer: six.