Find the absolute minimum value on \( (0, \infty) \) for \( f(x)=\frac{4 e^{x}}{x^{3}} \). Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute minimum is \( \square \) at \( x=\square \). (Round to two decimal places as needed.) B. There is no absolute minimum.

To find the absolute minimum value of \( f(x) = \frac{4e^x}{x^3} \) on the interval \( (0, \infty) \), we first need to find the critical points by taking the derivative and setting it to zero. 1. Compute \( f'(x) \): Using the quotient rule: \[ f'(x) = \frac{(x^3)(4e^x)'- (4e^x)(x^3)'}{(x^3)^2} \] Simplifying leads to: \[ f'(x) = \frac{4e^x (x^3 - 3x^2)}{x^6} = \frac{4e^x \cdot x^2 (x - 3)}{x^6}. \] Set \( f'(x) = 0 \) to find critical points: \[ 4e^x \cdot x^2 (x - 3) = 0. \] Since \( e^x > 0 \) and \( x^2 > 0 \) for \( x > 0 \), we have: \[ x - 3 = 0 \implies x = 3. \] 2. Evaluate the function at the critical point and the endpoints: \[ f(3) = \frac{4e^3}{3^3} = \frac{4e^3}{27}. \] Now check the behavior as \( x \to 0^+ \) and as \( x \to \infty \): - As \( x \to 0^+ \), \( f(x) \to \infty \). - As \( x \to \infty \), \( \( \frac{e^x}{x^3} \to \infty \). 3. Since the function tends to infinity in both directions and attains a minimum at \( x = 3 \), we can find \( f(3) \): \[ f(3) \approx \frac{4 \cdot 20.0855}{27} \approx 2.977. \] Thus, the absolute minimum value is: A. The absolute minimum is \( \approx 2.98 \) at \( x=3 \).