A function ƒ(x) describes a relationship between one or more inputs and one output. On the GRE, you can simply think of a function as an instruction for how to treat a particular variable or expression.

If ƒ(x) = 2x, then:

ƒ(2) = 4
ƒ(3) = 6

##### Example

Let ƒ(x) = 2x/4 – x

(a) What does ƒ(3) equal?

(b) What does ƒ(x + 1) equal?

### Solution

To solve, simply substitute the given value or expression into the function.

(a) Substitute 3 wherever you see x in the function.
ƒ(3) = 2(3)/4 – 3 = 6/1 = 6

(b) Substitute (x + 1) wherever you see x in the function.

ƒ(x + 1) = 2(x + 1)/4 – (x + 1) = 2x + 2/4 – x – 1 = 2x + 2/3 – x

#### Combining Functions

You can combine functions using any operation.

##### Example

Let ƒ(x) = 3x + 2 and g(x) = 4 – 5x

(a) Find ƒ(x) – g(x).

(b) Find ƒ(x) + g(x).

(b) Find ƒ(x) × g(x).

### Solution

(a) ƒ(x) – g(x) = (3x + 2) – (4 – 5x) = 3x + 2 – 4 + 5x = 8x – 2

(b) ƒ(x) + g(x) = (3x + 2) + (4 – 5x) = 3x + 2 + 4 – 5x = -2x + 6

(b) ƒ(x) × g(x) = (3x + 2) × (4 – 5x) = 12x – 15x2 + 8 – 10x = -15x2 + 2x + 8

#### Composite Functions

Another way of combining functions is with a composite function. This means the functions are nested, so you apply one function to find a value, then apply a second function to that value.

It is important to follow the order of operations, doing the inside function first, then the outside function.

Example

Let ƒ(x) = x2 and g(x) = x + 2

(a) Find ƒ(g(2)).

(b) Find g(ƒ(2)).

### Solution

(a) First evaluate g(2) since it is the inner function.
g(x) = x + 2, so g(2) = 2 + 2 = 4
Then apply ƒ(x) = x2
ƒ(4) = 42 = 16

(b) First evaluate ƒ(2) since it is the inner function.
ƒ(x) = x2, so ƒ(2) = 22 = 4
Then apply g(x) = x + 2
g(4) = 4 + 2 = 6

#### Variety of Symbols

On some questions, the functions won’t use the standard ƒ(x) or g(x) format. Instead, they will use symbols, including #, & and ♣.
The symbols can be confusing, but just treat them the same as any other function and these questions will be easy. Simply plug the numbers into the function.

Example

Let a # b = a + b.

(a) Find 2 # 3.

(b) Find (2 # 3) # 2.

### Solution

(a) You can think of this function as addition.
2 # 3 = 2 + 3 = 5

(b) First evaluate (2 # 3) since it is the inner function.
2 # 3 = 2 + 3 = 5
Then apply the same function, addition, again.
(2 # 3) # 2 = 5 # 2 = 7

##### Example

Let a @ b = ab and a & a = a2.

(a) Find a @ (a & b).

(b) Find (a @ a) & a.

### Solution

(a) First evaluate (a & b) since it is the inner function.
Since a & a = a2, you can think of this function as multiplication.
a & b = ab

Then apply a @ b = ab. Again, you can think of this function as multiplication.
a @ ab = a2b.

(b) First evaluate (a @ a) since it is the inner function.
Since a @ b = ab, you can think of this function as multiplication.
a @ a = a2

Then apply a & a = a2. Again, you can think of this function as multiplication.
a2 & a = a2 × a = a3

##### Example

For the numbers x, y, z, the function # is defined as x # y = xyx.

Find x # (y # z).

### Solution

Look for the function rule. Essentially, this function takes the first number, multiplies it by the second number, and then subtracts the first number.

First evaluate (y # z) since it is the inner function.
Since x # y = xyx, substitute y for x and z for y.
So y # z = yz − y.

Now substitute yzy into the function.
x # (y # z) = x # (yzy)

Apply the function rule:

• take the first number: x
• multiply it by the second number: (yzy)
• then subtract the first number: x

x # (yzy) = x × (yzy) − x = xyzxyx

Depending on the answer choices, you may need to factor out the x.
xyzxyx = x(yzy − 1)

Checking

One method to check your answer is to Plug In some numbers.

Choose some numbers for the variables. Let x = 1, y = 3 and z = 2.

x # (y # z) = 1 # (3 # 2)

Now use these numbers and apply the rule x # y = xyx.

Do the inner function first and evaluate (3 # 2).
Since x # y = xyx, substitute 3 for the first variable and 2 for the second variable.
x # y = xyx becomes (3)(2) − 3 = 3
So 1 # (3 # 2) becomes 1 # 3.

Since x # y = xyx, substitute 1 for the first variable and 3 for the second variable.
x # y = xyx becomes (1)(3) − 1 = 2

Check your answer by using the values x = 1, y = 3 and z = 2.
x # (y # z) = xyzxyx becomes (1)(2)(3) − (1)(3) − 1 = 6 − 3 − 1 = 2.

Since these values match the variables in the expression you found, you can infer that your expression is correct.

#### 800score Tip

When doing complex functions or algebra, plugging in numbers can help you better understand the equation and check your answer. However, this technique takes time, so use it judiciously.