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 $$ f^{\prime}(2)=0 $$ ?

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To find the value of $$ b $$ such that $$ f'(2) = 0 $$, we start by differentiating the function $$ f(x) $$:

$$
f(x) = (x^3 + bx + 6) g(x)
$$

We will use the product rule to differentiate $$ f(x) $$:

$$
f'(x) = (x^3 + bx + 6) g'(x) + (3x^2 + b) g(x)
$$

Next, we will evaluate $$ f'(2) $$:

1. Calculate $$ x^3 + bx + 6 $$ at $$ x = 2 $$:

$$
x^3 + bx + 6 = 2^3 + b \cdot 2 + 6 = 8 + 2b + 6 = 14 + 2b
$$

2. Calculate $$ g(2) $$:

Given $$ g(2) = 3 $$.

3. Calculate $$ g'(2) $$:

Given $$ g'(2) = -1 $$.

Now substituting these values into the expression for $$ f'(2) $$:

$$
f'(2) = (14 + 2b) g'(2) + (3 \cdot 2^2 + b) g(2)
$$

Calculating $$ 3 \cdot 2^2 + b $$:

$$
3 \cdot 2^2 + b = 3 \cdot 4 + b = 12 + b
$$

Now substituting $$ g(2) $$ and $$ g'(2) $$:

$$
f'(2) = (14 + 2b)(-1) + (12 + b)(3)
$$

Expanding this:

$$
f'(2) = - (14 + 2b) + (12 + b) \cdot 3
$$
$$
= -14 - 2b + 36 + 3b
$$
$$
= 22 + b
$$

We want $$ f'(2) = 0 $$:

$$
22 + b = 0
$$

Solving for $$ b $$:

$$
b = -22
$$

Thus, the value of $$ b $$ for which $$ f'(2) = 0 $$ is

$$
\boxed{-22}
$$
The value of $$ b $$ for which $$ f'(2) = 0 $$ is -22.

The function is given by , where is a
constant and is a differentiable function satisfying and
. For what value of is ?

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The function is given by , where is a constant and is a differentiable function satisfying and . For what value of is ?