Find the absolute minimum value on \( [0, \infty) \) for \( f(x)=(x+3)(x-3)^{2} \). Select the correct choice and, if necessary, fill in the answer boxes to complete your choice. A. The absolute minimum is \( \square \) at \( x=\square \). (Simplify your answers.) B. There is no absolute minimum

To find the absolute minimum value of \( f(x) = (x+3)(x-3)^2 \) on the interval \( [0, \infty) \), we first need to find the critical points by taking the derivative. 1. **Differentiate**: \[ f'(x) = (x+3)(2(x-3)) + (x-3)^2 = 2(x-3)(x+3) + (x-3)^2 \] Combine like terms: \[ f'(x) = (x-3)(2(x+3) + (x-3)) = (x-3)(3x + 3) \] 2. **Set the derivative to zero**: \[ (x-3)(3x + 3) = 0 \] This gives us two critical points: \[ x - 3 = 0 \implies x = 3 \] \[ 3x + 3 = 0 \implies x = -1 \quad (\text{not in } [0, \infty)) \] 3. **Evaluate \( f(x) \) at the critical point \( x=3 \)**: \[ f(3) = (3+3)(3-3)^2 = 6 \cdot 0 = 0 \] 4. **Evaluate \( f(x) \) at the endpoints of the interval**: - At \( x = 0 \): \[ f(0) = (0+3)(0-3)^2 = 3 \cdot 9 = 27 \] - As \( x \to \infty \), \( f(x) \to \infty \). Since \( f(3) = 0 \) is the lowest value we found compared to \( f(0) = 27 \) and as \( x \to \infty \), \( f(x) \) also goes to infinity. Thus, the absolute minimum is: A. The absolute minimum is \( 0 \) at \( x=3 \).