Functions can be daunting. Where do you plug in the number? Do you plug the equation in the function or the function in the equation? All of this might make you just want to unplug from the SAT altogether. But don’t despair.
Let’s do a quick introduction to functions before ending with a challenge question.
f(x) = x – 1
The above is a recipe, or instructions to follow once you are given the value of x inside the function. So we need an actual number to plug in:
f(2) = x – 1
The next step is to plug in ‘2’, wherever we see ‘x’, since the ‘2’ stands for ‘x’. Notice that it takes the place of ‘x’ in the original equation: f(x) = x – 1.
We can apply this to a more complicated scenario:
f(x) = x^2 – 3x, f(4) = 4^2 – 3(4) = 4
Where things get trickier is when there are multiple functions and/or when you make the function equal to something and you have to solve for some variable. Let’s deal with the first case: multiple functions.
f(x) = x – 2, g(x) = 3x
f(g(2)) can be translated as plugging the number ‘2’ inside the equation for g(x): 3(2) = 6. This leaves us with f(6). In other words g(2), which is on the very inside of the parentheses, is equal to 6.That still leaves the f function on the outer parentheses: f(6) = x – 2, so f(6) = 4.
Now, let’s make f(g(2) actually equal to something: f(g(2)) = y – 5. In this case, we want to find the value of ‘y’. We know that f(g(2)) is equal to 6. Therefore, 6 = y – 5, and y = 11.
The SAT will sometimes complicate things by making one function equal to another.
If f(x) = x^2 – 1, and g(x) = (f(x)/2) – 4, what is g(3)?
Here, just plug 3 into f(x), since we know that g(x) has a relationship to f(x): it’s 4 less than one half of f(x). f(3) = 3^2 – 1 = 8. 8/2 – 4 = 0.
If you got all that, then you have a good shot at the challenge problem below. Good luck!
Challenge Question
f(x) = √x– 1, and g(x) = x^2 – n. If f(g(3)) = 3, what is the value of g(n)?

(A) 7
(B) 1
(C) 16
(D) 49
(E) 56
Once you finish working out the problem on your own, watch the explanation video: