# SAT Newly Defined Functions

To solve **newly defined functions on the SAT**, you need to know how to

- apply the order of operations
- work with fractions
- manipulate algebraic expressions
- make simple substitutions
- apply the rules of exponents
- apply reasoning skills.

Some SAT questions use strange symbols that don't normally appear in a math problem. You will always be told what the symbols mean. For example, let \(a \oslash b=a^{b}\). The function \(a \oslash b\) is defined as the number to the left of the strange symbol, \(a\), raised to the power of \(b\), the number to the right of the strange symbol. So, \(3 \oslash 2\) will be equal to the number to the left of the strange symbol, \(3\), raised to the power of \(2\), the number to the right of the strange symbol. That is, \(3 \oslash 2=3^{2}\). The key to solving newly defined functions is to follow directions exactly.

## Examples

For any positive number \(x\), let \(x \Updownarrow \) be defined as \(x \Updownarrow = x^{2} + x\). What is the value of \(5\Updownarrow \)?

(A) \(\ \ 0\)

(B) \(\ \ 5\)

(C) \(\ \ 10\)

(D) \(\ \ 25\)

(E) \(\ \ 30\)

Correct Answer: E

Solution:

Tip: Follow directions exactly.

Start with the given definition and follow its instructions exactly:\[\begin{array}{l c l l l} x \Updownarrow &=& x^{2} + x &\qquad \text{given definition} &(1)\\ 5 \Updownarrow &=& 5^{2} + 5 &\qquad \text{plug values in definition} &(2)\\ &=& 25+5 &\qquad 5^{2} = 25 &(3)\\ &=& 30. &\qquad \text{add} &(4)\\ \end{array}\]

Incorrect Choices:

(A)

Tip: Eliminate obviously wrong answers.

We are told that \(x\) is positive. So, \(x^{2} + x\) can never equal 0.

(B)

Tip: Read the entire question carefully.

Tip: Follow directions exactly.

If you forget the first term in the definition and solve \(x \Updownarrow = x\) instead of the given \(x \Updownarrow = x^{2} + x\), you will get this wrong answer.

(C)

Tip: Read the entire question carefully.

Tip: Follow directions exactly.

If you forget to square the first term in the definition and solve \(x \Updownarrow = x + x\) instead of the given \(x \Updownarrow = x^{2} + x\), you will get this wrong answer.

(D)

Tip: Read the entire question carefully.

Tip: Follow directions exactly.

If you forget the second term in the definition and solve \(x \Updownarrow = x^{2}\) instead of the given \(x \Updownarrow = x^{2} + x\), you will get this wrong answer.

Let \(m\) and \(n\) be positive integers. If \( m \heartsuit n = (m-n)(m+n)\left(m^{2}+n^{2}\right)\), what is the value of \( ((0 \heartsuit 1) \heartsuit 0) \heartsuit 1\)?

(A) \(\ \ -1\)

(B) \(\ \ 0\)

(C) \(\ \ 1\)

(D) \(\ \ 2\)

(E) \(\ \ 3\)

Correct Answer: B

Solution 1:

Tip: Follow order of operations.

Start with the innermost parentheses \( (\underbrace{(0 \heartsuit 1)} \heartsuit 0) \heartsuit 1\).\[\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad \text{given definition} &(1)\\ 0 \heartsuit 1 &=& (0-1)(0+1)\left(0^{2}+1^{2}\right) &\qquad \text{follow directions exactly} &(2)\\ &=& (-1)(1)(1) &\qquad \text{simplify parentheses} &(3)\\ &=& -1. &\qquad \text{multiply} &(4)\\ \end{array}\]

Replacing \(0 \heartsuit 1\) in the original expression with \(-1\), we again evaluate the parentheses first: \( \underbrace{(-1 \heartsuit 0)} \heartsuit 1\).

\[\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right)&\qquad \text{given definition} &(5)\\ -1 \heartsuit 0 &=& (-1-0)(-1+0)\left((-1)^{2}+0^{2}\right)&\qquad \text{follow directions exactly} &(6)\\ &=& (-1)(-1)(1) &\qquad \text{simplify parentheses} &(7)\\ &=& 1. &\qquad \text{multiply} &(8) \end{array}\]

Replacing \( (-1 \heartsuit 0)\) in the previous expression with \(1\), we evaluate \(1 \heartsuit 1\).

\[\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad \text{given definition} &(9)\\ 1 \heartsuit 1 &=& (1-1)(1+1)\left((1)^{2}+1^{2}\right) &\qquad \text{follow directions exactly} &(10)\\ &=& (0)(2)(1) &\qquad \text{simplify parentheses} &(11)\\ &=& 0. &\qquad \text{multiply} &(12) \end{array}\]

Solution 2:

Tip: \(a^{2}-b^{2}=(a-b)(a+b)\)

Tip: Follow order of operations.

We simplify the definition of \(m \heartsuit n\) first, using the difference of squares formula.\[\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad (1)\\ &=& \left(m^{2}-n^{2}\right)\left(m^{2}+n^{2}\right) &\qquad (2)\\ &=& m^{4}-n^{4}. &\qquad (3)\\ \end{array}\]

Now when we apply \(m\heartsuit n\), we can use this simplified version of the definition:

\[\begin{array}{l c l l l} ((0 \heartsuit 1) \heartsuit 0) \heartsuit 1 &=& \left((0^{4}-1^{4}\right)\heartsuit 0) \heartsuit 1 &\qquad \text{apply}\ \heartsuit\ \text{on innermost parentheses} &(4)\\ &=& ((0-1)\heartsuit 0) \heartsuit 1 &\qquad \text{apply}\ 0^{a}=0\ \text{and}\ 1^{a}=1 &(5)\\ &=& (-1\heartsuit 0) \heartsuit 1 &\qquad \text{simplify innermost parentheses} &(6)\\ &=& \left((-1)^{4} - 0^{4}\right) \heartsuit 1 &\qquad \text{apply}\ \heartsuit\ \text{on parentheses} &(7)\\ &=& (1-0) \heartsuit 1 &\qquad (-1)^{4}=1\ \text{and}\ 0^{a}=0&(8)\\ &=& 1\heartsuit 1 &\qquad \text{simplify parentheses} &(9)\\ &=& 1^{4}-1^{4} &\qquad \text{apply}\ \heartsuit &(10)\\ &=& 1-1 &\qquad \text{apply}\ 1^{a}=1 &(11)\\ &=& 0. &\qquad \text{subtract} &(12) \end{array}\]

Incorrect Choices:

(A)

If in Solution 1 you stop at step \((4)\), only calculating \( (\underbrace{(0 \heartsuit 1)} \heartsuit 0) \heartsuit 1\), you will get this wrong answer.

(C)

If in Solution 1 you stop at step \((8)\), only calculating \( (\underbrace{(0 \heartsuit 1) \heartsuit 0)} \heartsuit 1\), you will get this wrong answer.

(D)

If you think the operator \(\heartsuit\) is a plus sign, you will get this wrong answer.

(E)

This wrong answer choice is just meant to confuse you.

For any positive integer \(x\) let \( \uplus\) be defined as \(y\uplus x = -x(-(-x^{y})-x)-x\). What is the value of \(2\uplus 3\)?

(A) \(\ \ -21\)

(B) \(\ \ -14\)

(C) \(\ \ -3\)

(D) \(\ \ 14\)

(E) \(\ \ 33\)

Correct Answer: A

Solution:

Tip: Follow directions exactly.

Tip: Follow order of operations.We have

\[\begin{array}{lclll} y\uplus x &=& -x\left(-\left(-x^{y}\right)-x\right)-x &\qquad \text{begin with the definition} &(1)\\ 2\uplus 3 &=& -3\left(-\left(-3^{2}\right)-3\right)-3 &\qquad \text{plug in}\ x\ \text{and}\ y &(2)\\ &=& -3(-(-9)-3)-3 &\qquad 3^{2}=9 &(3)\\ &=& -3(9-3)-3 &\qquad \text{use distributive property} &(4)\\ &=& -3(6)-3 &\qquad 9-3=6 &(5)\\ &=& -18-3 &\qquad \text{perform multiplication} &(6)\\ &=& -21. &\qquad \text{perform subtraction} &(7)\\ \end{array}\]

Incorrect Choices:

(B)

Tip: Follow directions exactly.

If you solve \(3\uplus 2 =-2\left(-\left(-2^{3}\right)-2\right)-2\), you will get this wrong answer. The question asks for \(2\uplus 3\), not for \(\boxed{3\uplus 2}\).

(C)

Tip: Follow order of operations.

If you forget to raise the second \(x\) to the power of \(y\), like this: \(-3(-(-\fbox{3})-3)-3=-3(3-3)-3=-3\), you will get this wrong answer.

(D)

This answer is offered to confuse you and make you want to choose between \(-14\) and \(14\).

(E)

Tip: Be careful with signs!

If in step \((3)\) you forget the negative sign in front of the squared term, you will get this wrong answer: \[\begin{array}{lclll} && -3(-\left(\fbox{-}3^{2}\right)-3)-3 & &(2)\\ &=& -3(-(9)-3)-3. &\qquad \text{mistake: forgot the negative sign} &(3)\\ \end{array}\]Alternatively, if in step \((4)\) you don't distribute the negative sign as shown below, you will get this wrong answer: \[\begin{array}{lclll} &&-3(\fbox{-}(-9)-3)-3 & &(3)\\ &=&-3(-9-3)-3. &\qquad \text{mistake: didn't distribute the negative sign} &(4)\\ \end{array}\]

## Review

If you thought these examples difficult and you need to review the material, these links will help:

- SAT Polynomials
- SAT Fractions and Decimals
- SAT Reasoning Skills
- Multi-Step Equations-Basic
- Applying the Difference of Two Squares Identity

\(a^{2}-b^{2}=(a+b)(a-b)\) - Applying the Perfect Square Identity

\((a \pm b)^{2}=a^{2} \pm 2ab + b^{2}\) - Rules of Exponents

\( a^m \times a^n = a^{ m + n } \), \( a^n / a^m = a^ { n - m }\ldots\)

## SAT Tips for Newly Defined Functions

- Follow directions exactly.
- Follow order of operations.
- Be careful with signs!
- SAT General Tips

**Cite as:**SAT Newly Defined Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-newly-defined-functions/