GRE Functions and How to Approach Them

If you’re feeling a little rusty on GRE algebra, you might not remember how to approach questions with algebraic functions. Don’t worry! We’ll go over all you need to know about GRE functions, with practice problems along the way.

What is a Function?

First things first: A function is like a “machine”. It is governed by the rules that dictate the output of each input. The specific rules that a function follows are given by the equation of the function.
For example, given the function \(f(x) = 6x – 7\) (read as “f of x”), we should know exactly how this particular “machine” works. It will always multiply the input by 6 and subtract 7 from that product. The result of the equation is its output.

Best Strategies for Solving GRE Functions

Very important note: Unlike in a simple equation format, in function notation, \(x\) is not just equal to a single value. Rather, \(x\) can be used to represent any value equal to any real number.

If you enter the input \(x = 2\) into this function, you will get the output of \(6(2) – 7 = 5\). We use the function notation to express this fact as \(f(2) = 5\), which shows that if you enter 2 into the function, you will get an output of 5.

Inputs Aren’t Always Numbers!

Earlier I said, “x can be used to represent any value equal to any real number” in a function. My past self didn’t exactly tell you the whole story. Sorry!

On the GRE, \(x\) can represent any real number OR any algebraic expression. In some problems, you may have to solve for something like \(f(k^2+2k+1)\).

Fear not! The basic structure of functions remains, even when dealing with algebraic expressions as inputs.

Just as \(f(2)\) equals \(6(2) – 7\), \(f(k^2 + 2k + 1)\) would equal \(6(k^2 + 2k + 1) – 7\). We’re still multiplying the input by 6 and subtracting 7 from that product.

Function-ception?

But that’s still not the whole story! Problems on the GRE can even have functions… within functions. If we have, for example, \(f(g(x))\), then this would be read as “f of g of x”, and we would input some value into the function \(g\), and take the output from \(g\) and use it as the input for \(f\).

Broken down, if \(f(x) = 2x + 1\) and \(g(x) = 3x + 2\), and we’re asked to solve for \(f(g(3))\):

3 first enters the \(g\) function, giving us the output \(g(3) = 3(3) + 2 = 11\).

\(g\)‘s output now becomes our input for the \(f\) function: \(f(11) = 2(11) +1 = 23\)

We are essentially substituting the equation of \(g(x)\) in for the variable in the equation for \(f(x)\).
\(f(g(x)) = f(3x + 2) = 2(3x +2) + 1 = 6x + 5\)
\(f(g(3)) = 6(3) + 5 = 23\)

Try your hand at this problem featuring functions within functions!

Understand the Parentheses

Remember that the parentheses in the function notation are mathematically inviolable. This means that you cannot perform any mathematical operations on the contents of the parentheses. You cannot distribute, factor out, etc.

\(
\mathrm{f}(\mathrm{x}+5) \neq \mathrm{f}(\mathrm{x})+5\)
\(\mathrm{f}(\mathrm{x}-5) \neq \mathrm{f}(\mathrm{x})-5\)
\(\mathrm{f}\left(5^{*} \mathrm{x}\right) \neq 5^{*} \mathrm{f}(\mathrm{x})\)
\(f\left(\frac{x}{5}\right) \neq \frac{f(x)}{5}\)
\(\mathrm{f}\left(\mathrm{x}^{2}\right) \neq[\mathrm{f}(\mathrm{x})]^{2}
\)
Imagine the “x” in a function to be a blank space where you can put any sort of number or algebraic expression. Anything that gets plugged into the space on the left (in the parentheses) needs to get plugged into the space(s) on the right.

Simplify wherever possible as you go along your solution process, especially if you’re dealing with complicated expressions and/or functions within functions. If you don’t simplify as you solve, you may end up with a big mathematical mess.

With these strategies in mind, you should be able to approach GRE functions with more confidence now. Good luck!