Find a) any critical values and b) any relative extrema.
\[
g(x)=-x^{3}+12 x-20
\]

Step 1 ：Given the function \(g(x) = -x^{3} + 12x - 20\)

Step 2 ：Find the derivative of the function \(g'(x) = -3x^{2} + 12\)

Step 3 ：Set the derivative equal to zero and solve for x to find the critical values: \(-3x^{2} + 12 = 0\) gives \(x = -2, 2\)

Step 4 ：Find the second derivative of the function \(g''(x) = -6x\)

Step 5 ：Substitute the critical values into the second derivative: \(g''(-2) = 12\) and \(g''(2) = -12\)

Step 6 ：Since \(g''(-2) > 0\), -2 is a relative minimum. Since \(g''(2) < 0\), 2 is a relative maximum

Step 7 ：Final Answer: The critical values are \(\boxed{-2}\) and \(\boxed{2}\). The relative minimum is at \(\boxed{-2}\) and the relative maximum is at \(\boxed{2}\)