The function \( f \) is given by \( f(x)=\left(x^{3}+b x+6\right) g(x) \), where \( b \) is a constant and \( g \) is a differentiable function satisfying \( g(2)=3 \) and \( g^{\prime}(2)=-1 \). For what value of \( b \) is \( f^{\prime}(2)=0 \) ?

To find the value of \( b \) such that \( f^{\prime}(2) = 0 \), we first need to differentiate the function \( f(x) = (x^3 + b x + 6) g(x) \) using the product rule. The product rule states that if \( h(x) = u(x)v(x) \), then \( h'(x) = u'(x)v(x) + u(x)v'(x) \). Here, we can set \( u(x) = x^3 + bx + 6 \) and \( v(x) = g(x) \). First, we calculate the derivatives: 1. The derivative of \( u(x) \) is: \[ u'(x) = 3x^2 + b \] 2. The derivative of \( v(x) \) is \( g'(x) \). Now we apply the product rule to differentiate \( f(x) \): \[ f'(x) = u'(x)g(x) + u(x)g'(x) = (3x^2 + b)g(x) + (x^3 + bx + 6)g'(x) \] Next, we evaluate \( f'(2) \): \[ f'(2) = (3(2^2) + b)g(2) + (2^3 + b(2) + 6)g'(2) \] \[ = (3(4) + b)g(2) + (8 + 2b + 6)g'(2) \] \[ = (12 + b)g(2) + (14 + 2b)g'(2) \] Now plug in the known values \( g(2) = 3 \) and \( g'(2) = -1 \): \[ f'(2) = (12 + b)(3) + (14 + 2b)(-1) \] \[ = 36 + 3b - 14 - 2b \] \[ = 36 - 14 + (3b - 2b) \] \[ = 22 + b \] We want \( f'(2) = 0 \): \[ 22 + b = 0 \] \[ b = -22 \] So, the value of \( b \) for which \( f'(2) = 0 \) is \( \boxed{-22} \).