The function $$ f $$ and $$ g $$ are such that $$ f(x)=3 x-1 $$ and $$ g(x) $$ $$ =x^{2}+4 $$. Given that $$ f g(x)=2 g f(x) $$, what is the value of $$ 15 x^{2}-12 x-1 ? $$ Select one: a. -1 b. $$ \frac{1}{2} $$ c. 0 d. 1

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Given the functions
$$
f(x)=3x-1 \quad \text{and} \quad g(x)=x^2+4,
$$
we have the equation
$$
f\circ g(x)=2\,(g\circ f(x)).
$$

First, compute $$ f\circ g(x) $$:
$$
f(g(x)) = 3\,(g(x))-1 = 3\,(x^2+4)-1 = 3x^2+12-1 = 3x^2+11.
$$

Next, compute $$ g\circ f(x) $$:
$$
g(f(x)) = \bigl( f(x) \bigr)^2+4 = (3x-1)^2+4.
$$
Expanding the square:
$$
(3x-1)^2 = 9x^2-6x+1,
$$
so
$$
g(f(x)) = 9x^2-6x+1+4 = 9x^2-6x+5.
$$

Now set up the given equation:
$$
3x^2+11 = 2\,(9x^2-6x+5).
$$
Simplify the right-hand side:
$$
3x^2+11 = 18x^2-12x+10.
$$

Rearrange the equation by subtracting $$ 18x^2-12x+10 $$ from both sides:
$$
3x^2+11 -18x^2+12x-10 = 0.
$$

Combine like terms:
$$
-15x^2+12x+1 = 0.
$$
Multiply the whole equation by $$-1$$:
$$
15x^2-12x-1 = 0.
$$

The expression $$ 15x^2-12x-1 $$ is equal to $$ 0 $$ for the value(s) of $$ x $$ that satisfy the given condition. Therefore, the value is:
$$
\boxed{0}.
$$
The value of $$ 15x^{2}-12x-1 $$ is 0.

The function \( f \) and \( g \) are such that \( f(x)=3 x-1 \) and \( g(x) \) \( =x^{2}+4 \). Given that \( f g(x)=2 g f(x) \), what is the value of \( 15 x^{2}-12 x-1 ? \) Select one: a. -1 b. \( \frac{1}{2} \) c. 0 d. 1

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