Find the linearization $L(x)$ of the function $g(x)=x f\left(x^{2}\right)$ at $x=2$ given the following information.
\[
f(2)=-1 \quad f^{\prime}(2)=12 \quad f(4)=6 \quad f^{\prime}(4)=-4
\]

Answer: $L(x)=$

Step 1 ：Given the function \(g(x) = x f(x^2)\), we need to find the linearization at \(x=2\).

Step 2 ：The linearization of a function \(g(x)\) at a point \(x=a\) is given by the formula: \(L(x) = g(a) + g'(a)(x-a)\).

Step 3 ：First, we find \(g(2)\). Substituting \(x=2\) into \(g(x)\), we get \(g(2) = 2f(2^2) = 2f(4) = 2*6 = 12\).

Step 4 ：Next, we find \(g'(x)\), the derivative of \(g(x)\). Using the chain rule and product rule, we get: \(g'(x) = f(x^2) + 2x^2 f'(x^2)\).

Step 5 ：Substituting \(x=2\) into \(g'(x)\), we get \(g'(2) = f(2^2) + 2*2^2*f'(2^2) = f(4) + 8f'(4) = 6 + 8*(-4) = -26\).

Step 6 ：Now, we can find the linearization \(L(x)\) at \(x=2\): \(L(x) = g(2) + g'(2)(x-2) = 12 - 26(x-2) = 12 - 26x + 52 = -26x + 64\).

Step 7 ：So, the linearization \(L(x)\) of the function \(g(x)=x f\left(x^{2}\right)\) at \(x=2\) is \(\boxed{L(x) = -26x + 64}\).