Find the maximum or minimum value of the function g(x): $g(x)=-3 x^{2}+18 x-29$

Step 1 ：Graph the function or try different values of \(x\) to find that the function seems to be maximized at \(x = 3\), which would make the maximum value 2.

Step 2 ：Consider the expression \(2 - g(x) = 3x^2 - 18x + 31\). This is zero at \(x = 3\), so \(x - 3\) is a factor: \(2 - g(x) = (x - 3)^2 (3x - 31)\).

Step 3 ：Since \(x\) is a real number, \(3x - 31\) can be both positive and negative. Hence, \(g(x) \le 2\) for all \(x\), which confirms that the maximum value is \(\boxed{2}\).

Step 4 ：For the minimum value, find the roots of the derivative of \(g(x)\), which is \(g'(x) = -6x + 18\). Setting this equal to zero gives \(x = 3\).

Step 5 ：Substitute \(x = 3\) into \(g(x)\) to get \(g(3) = -3(3)^2 + 18(3) - 29 = -27 + 54 - 29 = -2\). Hence, the minimum value of \(g(x)\) is \(\boxed{-2}\).