{"id":1893,"date":"2019-09-24T16:58:40","date_gmt":"2019-09-24T21:58:40","guid":{"rendered":"http:\/\/www.kaptest.com\/blog\/prep\/?p=1893"},"modified":"2021-09-13T16:29:22","modified_gmt":"2021-09-13T16:29:22","slug":"psat-math-functions","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/psat\/psat-math-functions\/","title":{"rendered":"PSAT Math: Functions"},"content":{"rendered":"

Functions act as rules that transform inputs into outputs, and they differ from equations in that each input must have only one corresponding output. For example, imagine a robot: Every time you give it an apple, it promptly cuts that apple into three slices.<\/p>\n

The following table summarizes the first few inputs and their corresponding outputs.<\/p>\n\n\n\n\n\n\n\n
Domain,\u00a0x<\/i>: # apples given to robot<\/b><\/td>\nRange,\u00a0f<\/i><\/b>\u200a<\/b>(x<\/i>): # slices returned by robot<\/b><\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n3<\/td>\n<\/tr>\n
2<\/td>\n6<\/td>\n<\/tr>\n
3<\/td>\n9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the table you see that the output will always be triple the input, and you can express that relationship as the function f<\/i>\u200a(x<\/i>) = 3x<\/i> (read \u201cf<\/i> of x<\/i> equals three x<\/i>\u201d).<\/p>\n

PSAT questions, especially those involving real-world situations, might ask you to derive the equation of a function, so you\u2019ll need to be familiar with the standard forms. Following is the standard form of a linear function:<\/p>\n

f<\/i>\u200a(x<\/i>) = kx<\/i> + f<\/i>\u200a(0)<\/p>\n

The input, or domain<\/b>, is the value represented by x<\/i>. Sometimes the domain will be constrained by the question (e.g., x<\/i> must be an integer). Other times, the domain could be defined by real-world conditions.<\/p>\n

For example, if x<\/i> represents the time elapsed since the start of a race, the domain would need to exclude negative numbers.<\/p>\n

The output, or range<\/b>, is what results from substituting a domain value into the function and is represented by f<\/i>(x<\/i>).<\/p>\n

The initial amount, or y<\/i>-intercept<\/b>, is represented by f<\/i>(0)\u2014the value of the function at the very beginning.<\/p>\n

If you think this looks familiar, you\u2019re absolutely right. It\u2019s just a dressed-up version of the standard y<\/i> = mx<\/i> + b<\/i> equation you\u2019ve already seen.<\/p>\n

Take a look at the following table for a translation:<\/p>\n\n\n\n\n\n\n
Function Notation<\/b><\/td>\nWhat It Represents<\/b><\/td>\nSlope-Intercept Counterpart<\/b><\/td>\n<\/tr>\n
f<\/i>(x<\/i>)<\/td>\ndependent variable or output<\/td>\ny<\/i><\/td>\n<\/tr>\n
k<\/i><\/td>\nrate of change, slope<\/td>\nm<\/i><\/td>\n<\/tr>\n
f<\/i>(0)<\/td>\ny<\/i>-intercept or initial quantity in a word problem<\/td>\nb<\/i><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

As you might have guessed, an exponential equation has a standard function notation as well. Here we\u2019ve used g<\/i> in place of f<\/i> for visual clarity. Know that the letter used to represent a function (f, g, h,<\/i> etc.) is sometimes arbitrarily chosen.
\ng<\/i>(x<\/i>) = g<\/i>(0)(1+ r<\/i>)x<\/sup><\/i><\/p>\n

Just as before, g<\/i>(0) represents the initial amount and r<\/i> represents the growth (or decay) rate. Recognizing that function notation is a variation of something you already know will go a long way toward reducing nerves on Test Day. You should also note that graphing functions is a straightforward process: In the examples above, just replace f<\/i>(x<\/i>) or g<\/i>(x<\/i>) with\u00a0y<\/i> and enter into your graphing calculator.<\/p>\n

Note<\/b><\/p>\n

A quick way to determine whether an equation is a function is to conduct the vertical line test: If a vertical line passes through the graph of the equation more than once for any given value of x<\/i>, the equation is not a function.<\/p>\n

Below\u00a0is an example of a test-like question about functions.<\/p>\n

\t
\t\t
<\/p>\n
1. If\u00a0f<\/i>\u200a(x<\/i>) =\u00a0x<\/i>\u200a\u200a2<\/sup>\u00a0\u2013\u00a0x<\/i>\u00a0for all\u00a0x<\/i>\u00a0\u2264 \u20131<\/span>\u00a0and\u00a0f<\/i>\u200a(x<\/i>) = 0<\/span>\u00a0for all\u00a0x<\/i>\u00a0> \u20131, which of the following could not be a value\u00a0of\u00a0f<\/i>\u200a(x<\/i>) ?<\/span><\/section>\n

A) \u2013\u200a4
\nB) 0
\nC)\u00a02
\nD) 4<\/p>\n<\/div><\/div><\/p>\n<\/section>\n

Use the Kaplan Method for Math to solve this question, working through it step-by-step.<\/p>\n

The following table shows Kaplan\u2019s strategic thinking on the left, along with suggested math scratchwork on the right.
\n

STRATEGIC THINKING<\/strong><\/td>MATH SCRATCHWORK<\/strong><\/td><\/tr>
Step 1: Read the question, identifying and organizing important information as you go<\/p>\n

The question is asking for the answer choice that could not be in the range of this function.<\/td>

<\/td><\/tr>

\nStep 2: Choose the best strategy to answer the question<\/p>\n

How can you determine which of the four choices is not part of the range?<\/p>\n

You\u2019ll need to examine each piece of the domain individually to learn more.<\/p>\n

First, summarize each piece of the domain description.<\/p>\n

Can you use the domain information to eliminate any answer choices?<\/p>\n

What happens when x > \u22121?<\/p>\n

Because the range is 0 when x > \u22121, you can eliminate B.<\/p>\n

Would your knowledge of number properties be helpful for the second domain component?<\/p>\n

Absolutely. Because the range x\u00b2 \u2212 x only applies when x \u2264 \u22121, you only need to consider what happens when a negative number less than \u22121 is substituted for x.<\/p>\n

No matter what number you use, your output will always be positive. Eliminate C and D.<\/td>

domain x \u2264 \u22121: range = x\u00b2 \u2212 x domain x > \u22121: range = 0 (negative)\u00b2 \u2212 negative \u2192 positive + positive \u2192 positive<\/td><\/tr>
Step 3: Check that you answered the right question<\/p>\n

The range of f(x) consists of only positive numbers, so (A) is correct.<\/td>

\u22124 is not positive<\/td><\/tr><\/tbody><\/table><\/div>