Explanation

Plugging 10 values into two compound functions is going to involve lots of arithmetic and will take a long time, so it is better to do this one algebraically.

Working from the inside out, find Quantity A: \(f(g(x))=f(x+1)=3(x+1)^2=3x^2+6x+3\); remember to FOIL the (x + 1) when you square it. Similarly, find Quantity B: \(g(f(x)) = g(3x^2) = 3x^2 + 1\).

You can add or subtract the same value from both quantities without affecting which is bigger; doing so with \(3x^2 + 1\) leaves you with 6x + 2 in Quantity A and 0 in Quantity B.

Because 6x + 2 is a linear function whose graph is a line with positive slope, you know that the values of the function will increase as the values of x increase.

So you only need to plug in the endpoints of the given range of x-values to see what happens to the function: 6(–10) + 2 = –58 and 6(–1) + 2 = –4.

So all possible values of Quantity A are still less than 0, and the answer is choice (B).