SAT Tip: Putting Algebra Skills to the Test
This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.
Function questions on the SAT can be intimidating to students who have never seen them before, especially if they come with the dreaded “f(x)” symbol.
Remember that we can think of “f(x)” as simply a replacement for “y.” Rather than write an equation y = 5x + 3, for example, we can write it as f(x) = 5x + 3 and call it a “function.” We can also write functions with other variables: f(m) = 5m + 3 is the same function as f(x) = 5x + 3. The “x” or “m” is a placeholder for the domain, or what goes into the function. The f(x) or f(m) is a placeholder for the range, or what comes out of the function. It’s helpful to think of functions like machines. A value goes in one end, and a value comes out the other end.
A simple value question on the SAT might look something like this:
What is the value of f(5) for the function f(z) = z2 – 9?
To solve, we simply rewrite the function, plugging 5 in for z:
f(5) = (5)2 – 9
f(5) = 25 – 9
f(5) = 16
Harder function question may require you to plug in values and rewrite functions more than once. I call these “double-functions.” For example:
What is the value of f(f(5)) for the same equation?
The symbol “f(f(5))” means that we’ll first plug in z = 5, just as we did above. When we get our answer of 16, we’re still left with f(16), so we’ll have to plug in z = 16 and simplify the equation again.
f(16) = (16)2 – 9
f(16) = 256 – 9
f(16) = 247
OK, but what about those tricky questions that put numbers outside the parentheses and in front of the function symbol? These questions might ask something like, what is the value of 2f(z)?
Since the number is outside the parentheses, we’re being told to multiply the entire equation by 2:
f(z) = z2 – 9
2f(z) = 2(z2 – 9)
Using the property of distribution, let’s simplify:
2f(z) = 2z2 – 18
Now try a question on your own:
If f(x) = 3x – 4, what is the value of q if 2f(q) = f(3q)?
Stumped? To begin with, remember that the “x” is just a placeholder for the domain, so we can rewrite the function in terms of “q” or any other variable:
f(q) = 3q – 4
We’re asked to find “q” if 2f(q) = f(3q), so first we must find the value of 2f(q), then the value of f(3q):
2f(q) = 2(3q – 4)
2f(q) = 6q – 12
Now for f(3q), we’ll just plug in “3q” in for “q”:
f(3q) = 3(3q) – 4
f(3q) = 9q – 4
Because the question stem tells us that 2f(q) = f(3q), we know that 6q – 12 = 9q – 4, using substitution. Now we can find q:
6q – 12 = 9q – 4
-12 = 3q – 4
-8 = 3q
-8/3 = q
Remember that 90 percent of the algebra with functions is easy. All we have to do is know where to plug in, and where to substitute.
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